1.﹣22=()

A.﹣2B.﹣4C.2D.4

The analysis is based on the power algorithm.

Solution: -22 =-4

So choose B.

This review examines power, and the key to solving this problem is to master the algorithm of power.

2. The average distance between the sun and the earth is about 150000000 kilometers, and the data of 15000000 is expressed as () by scientific notation.

a . 1.5× 108 b . 1.5× 109 c . 0. 15× 109d

The expression of analytical scientific notation is a× 10n, where 1 ≤| a | < 10, and n is an integer. When determining the value of n, it depends on how many digits the decimal point has moved when the original number becomes a, and the absolute value of n is the same as that of the decimal point. When the absolute value of the original number is greater than 1, and when the absolute value of the original number is less than 1, n is negative.

Solution: 150000000 is expressed in scientific notation as: 1.5× 108.

So choose a.

Comment on this topic and investigate the expression method of scientific notation. The expression of scientific notation is a× 10n, where 1 ≤| A | < 10, and n is an integer. It is very important to correctly determine the value of a and the value of n when expressing.

3. As shown in the figure, in △ABC, point D and point E are on the AB side and AC side, respectively, DE∨BC. If BD=2AD, then ().

Asian Development Bank.

Find △ADE∽△ABC according to the meaning of the question, and then use the known to find the ratio of the corresponding sides.

Solution: ∫DE∨BC,

∴△ADE∽△ABC，

BD = 2AD，

∴===,

Then =,

∴A, C and D options are wrong, and B option is correct.

Therefore, choose: B.

This review mainly examines similar triangles's judgment and nature, and correctly drawing the proportion of the corresponding side is the key to solving the problem.

4.| 1+|+| 1﹣|=()

A. 1B。 C.2D.2

According to the nature of absolute value, we can get the answer.

Solution: The original formula 1++- 1 = 2,

Therefore, choose: d.

This question examines the properties of real numbers, and using the absolute value of difference as a large number to reduce the number is the key to solving the problem.

5. Let x, y and c be real numbers, ()

A. if x=y, then x+c = y-CB. If x=y, then xc=yc.

C. if x=y, then d. if x = 3y, then 2x=3y.

According to the properties of the equation, we can get the answer.

Solution: a, add different numbers on both sides, so a does not meet the meaning of the question;

B, both sides are multiplied by c, so b meets the meaning of the question;

When c and c=0, it is meaningless to divide both sides by c, so c does not meet the meaning of the question;

D, the two sides are multiplied by different numbers, so d does not meet the meaning of the question;

Therefore, choose: B.

This question reviews the nature of the equation, and memorizing the nature of the equation and solving it according to the nature of the equation is the key to solving the problem.

6. if x+5 > 0, then ()

A.x+ 1 < 6, so this option does not meet the meaning of the question;

So choose C.

In this paper, the properties of inequality are investigated, and correct deformation according to the properties of inequality is the key to solve this problem.

7. The number of tourists in a scenic spot is increasing year by year. According to statistics, it was1080,000 in 2065.438+06 and1680,000 in 2065.438+06. Assuming that the average annual growth rate of visitors is X, then ()

a. 10.8( 1+x)= 16.8b. 16.8( 1﹣x)= 10.8

c . 10.8( 1+x)2 = 16.8d . 10.8[( 1+x)+( 1+x)2]= 16.8

Suppose the average annual growth rate of the number of visitors is X, and according to the meaning of the question, we can get the equivalent relationship:108,000 person-times ×( 1+ growth rate) 2 =168,000 person-times. Just list the equations according to the equivalence relation.

Solution: Let the average annual growth rate of the number of visitors be x, which is derived from the meaning of the question:

10.8( 1+x)2 = 16.8，

So choose: C.

This topic mainly examines the quadratic equation of one variable abstracted from practical problems. If the quantity before the change is A, the quantity after the change is B and the average change rate is X, then the quantity relationship after the two changes is A (1x) 2 = B. 。

8. As shown in the figure, in Rt△ABC, ∠ ABC = 90, AB=2, BC = 1. Rotate △ABC around straight lines AB and BC respectively, and the circumference of the ground circle of the obtained geometric figure is marked as l 1, and l2 and lateral area are marked as S 1 and S2 respectively, then

A.l 1:l2= 1:2，S 1:S2= 1:2b . l 1:L2 = 1:4，s 1:S2 = 1:2

C.l 1:l2= 1:2，S 1:S2= 1:4d . l 1:L2 = 1:4，s 1:S2 = 1:4

Analysis: Calculate l 1, l2 according to the circumference of the circle, and then calculate S 1, S2 according to the area formula of the sector to get the ratio.

Solution: ∫l 1 = 2π×BC = 2π,

l2=2π×AB=4π，

∴l 1:l2= 1:2，

∵S 1=×2π×=π，

S2 =×4π×2π，

∴S 1:S2= 1:2，

So choose a.

This question examines the calculation of a cone, mainly using the circumference of a circle as 2πr and lateral area =lr as the key to solving the problem.

9. Let the straight line x= 1 be the image symmetry axis of the function y=ax2+bx+c(a, b and c are real numbers and A < 0), ().

A. if m > 1, then (m > 1) a+b > 0b. If m > 1, then (m > 1) a+b < 0.

C. if m < 1, then (m < 1) a+b > 0d. If m < 1, then (m < 1) a+b < 0.

B =-2a can be obtained according to the symmetry axis, and the answer can be obtained according to the rational number multiplication.

Solution: From the symmetry axis, we can get

b=﹣2a.

(m﹣ 1)a+b=ma﹣a﹣2a=(m﹣3)a

(m < 3) a > 0 when m < 1,

So choose: C.

In this paper, the relationship between quadratic function image and coefficient is investigated, and it is concluded that b =-2a is the key to solve the problem by using symmetry axis.

10. As shown in the figure, in △ABC, AB=AC, BC= 12, E is the midpoint of the AC side, and the intersection BC of the midline of the line segment BE is at point D. Let BD=x, tan∠ACB=y, then ().

a.x﹣y2=3b.2x﹣y2=9c.3x﹣y2= 15d.4x﹣y2=2 1

After analysis, A is in Q, E is in M, and DE is connected. According to the vertical line in the line segment, the right triangle is solved to get DE=BD=x, BD=DC=6, CM=DM=3, EM=3y, AQ=6y. In Rt△DEM, it can be obtained according to Pythagorean theorem.

Solution:

A is AQ⊥BC in Q, E is EM⊥BC in M, connecting DE,

∵ the middle vertical line of be passes through BC to D, BD=x,

∴BD=DE=x，

AB = AC，BC= 12，tan∠ACB=y

∴==y,BQ=CQ=6，

∴AQ=6y，

∵AQ⊥BC,EM⊥BC，

∴AQ∥EM，

E is the midpoint of communication,

∴CM=QM=CQ=3，

∴EM=3y，

∴DM= 12﹣3﹣x=9﹣x，

In Rt△EDM, according to Pythagorean theorem, x2 = (3y) 2+(9-x) 2,

That is 2x-y2 = 9,

So choose B.

Comment on this question, we have investigated the properties of the midline of the line segment, the properties of the isosceles triangle, the pythagorean theorem, the solution of the right triangle and other knowledge points. Making auxiliary lines correctly is the key to solve this problem.

Step 2 fill in the blanks

1 1. The median of data 2, 2, 3, 4 and 5 is 3.

According to the definition of median, the data should be arranged from small to large, and the middle number (or the average of two numbers) is the median, so you can get the answer.

Solution: from small to large, the order is: 2, 2, 3, 4, 5,

The middle number is 3,

Then the median of this set of numbers is 3.

So the answer is: 3.

Commenting on this question examines the median. Pay attention to the order when looking for the median, and then determine the median according to odd and even numbers. If there are odd numbers in the data, the middle number is what you want. If there are even numbers, find the average of the middle two digits.

12. As shown in the figure, AT⊙O cuts to point A, and AB is the diameter of ⊙O. If ∠ ABT = 40, then ∠ ATB = 50.

Analysis can find the answer according to the nature of tangent.

Solution: ∵ the incision at point A ⊙O, AB is the diameter ⊙O,

∴∠BAT=90，

∫∠ABT = 40，

∴∠ATB=50，

So the answer is: 50.

The key to solve this problem is to find ∠ ATB = 90 according to the nature of tangent, which belongs to the basic question type.

13. There are three balls (only in different colors) in an opaque cloth bag, two of which are red balls, and 1 is white balls. Pick a ball at random, write down the color, put it back, stir it evenly, and the probability of picking another ball at random is.

Draw the corresponding tree diagram according to the meaning of the question, find out the number of all possible situations, and then find out the number of situations that are red balls twice, so as to find out the probability.

Solution: Draw the corresponding tree diagram according to the meaning of the question.

So there are nine situations for a * * *, and four situations for a red ball to touch twice.

The probability that both are red balls is,

So the answer is:

The key to solve this problem is to comment on this question, list method and tree diagram examination, and draw the corresponding tree diagram according to the meaning of the question.

14. If |m|=, then m=3 or | 1.

By analyzing the properties of absolute value and fraction, we can get m﹣ 1≠0, m﹣3=0 or |m|= 1, and we can get m 。

Solution: Judging from the meaning of the problem,

m﹣ 1≠0，

Then m≠ 1,

(m﹣3)|m|=m﹣3，

∴(m﹣3)(|m|﹣ 1)=0，

∴m=3 or m = 1,

∵m≠ 1，

∴m=3 or m =1,

So the answer is: 3 or-1

The comment on this question mainly examines the nature of absolute value and fraction, and memorizing that the denominator of fraction is not 0 is the key to solve this problem.

15. As shown in the figure, in Rt△ABC, ∠ BAC = 90, AB= 15, AC=20, D is on the side of AC, AD=5, DE⊥BC is on E, AE is connected, so the area of △ABE is equal to 78.

Prove the area of BC==25, △ABC = 150 by Pythagorean theorem, and prove that △CDE∽△CBA, CE= 12, and BE=BC﹣CE= 13. Then we can get the answer from the area relationship of triangle.

Solution: ∵ in Rt△ABC, ∠ BAC = 90, AB= 15, AC=20,

∴BC==25, area △ ABC = ABAC =×15× 20 =150,

AD = 5，

∴CD=AC﹣AD= 15，

∵DE⊥BC，

∴∠DEC=∠BAC=90，

And ≈C =∠C,

∴△CDE∽△CBA，

That is,

Solution: CE= 12,

∴BE=BC﹣CE= 13，

∫△ABE area :△△ ABC area = Be: BC = 13: 25,

∴△ABE area =×150 = 78;

So the answer is: 78.

Comment on this topic to examine similar triangles's judgment and nature, Pythagorean theorem, triangle area; Mastering Pythagorean Theorem skillfully and proving triangle similarity is the key to solve the problem.

16. A fruit shop sells 50 kilograms of bananas. On the first day, the price was 9 yuan/Jin. The next day, the price was reduced by 6 yuan/Jin. On the third day, it dropped to 3 yuan/Jin. Sold for three days, all sold out, 270 yuan. If the store sells t kilograms of bananas the next day, it will sell 30 kilograms of bananas the next day, which is calculated according to the sales volume for three days.

Solution: If you sell X kilograms of bananas on the third day, you will sell (50-t-x) kilograms of bananas on the first day.

According to the question, it means: 9 (50-t-x)+6t+3x = 270,

Then x = = 30,

So the answer is: 30.

This topic review mainly examines algebraic expression ability. The key to solving the problem is to understand the meaning of the problem and list the equations by grasping the equation relationship, thus representing the kilograms of bananas sold on the third day.

Three. solve problems

17. In order to know the high jump level of ninth-grade students in a school, 50 students of this grade were randomly selected for high jump test, and the test results were drawn into a frequency table and an unfinished frequency histogram as shown in the figure (each group contains the front boundary value, but does not include the back boundary value).

Frequency table of high jump scores of 50 ninth-grade students in a school

Frequency of group (m)

1.09~ 1. 198

1. 19~ 1.29 12

1.29~ 1.39A

1.39~ 1.49 10

(1) Find the value of a and complete the frequency histogram;

There are 500 students in this grade. Estimated number of students with high jump scores of 1.29m (inclusive) in this grade.

Analyze (1) the total number of people 50 MINUS the number of people in other groups to get a value;

(2) The total number of people can be multiplied by the corresponding proportion.

Solution: (1) A = 50-8-12-10 = 20,

;

(2) The number of people with high jump scores above 1.29m (including 1.29m) in this grade is 500×300 (people).

This topic examines the ability to read the histogram of frequency distribution and the ability to obtain information by using statistical graphs. When using statistical charts to obtain information, we must carefully observe, analyze and study statistical charts in order to make correct judgments and solve problems. It also checked the overall sample estimate.

18. In the plane rectangular coordinate system, the linear function y=kx+b(k, b is constant, and the image of k≠0 passes through points (1 0) and (0,2).

(1) When < 2 < x ≤ 3, find the value range of y;

(2) Given a given point P(m, n) on the image of this function, and m-n = 4, find the coordinates of point P. 。

This analysis can be obtained by finding a resolution function by undetermined coefficient method;

(1) can be obtained by adding or subtracting linear functions.

(2) According to the meaning of the question, n =-2m+2 is obtained, and the equation can be solved simultaneously.

Solution: let the analytical formula be: y=kx+b,

Substituting (1, 0) and (0,2) gives:,

Solution:

The analytical formula of this function is: y =-2x+2;

(1) Substitute X =-2 for Y =-2x+2, and y=6.

Substitute x=3 into y =-4 y =-2x+2,

The range of ∴y is ∴ 4 ≤ y < 6.

(2)∵ point P(m, n) is on the image of this function,

∴n=﹣2m+2，

∵m﹣n=4，

∴m﹣(﹣2m+2)=4，

The solution is m=2, n =-2,

The coordinate of point P is (2, -2).

Comment on this topic, we have investigated the analytical formula of solving the linear function by using the undetermined coefficient method, the coordinate characteristics of each point on the image of the linear function and the properties of the linear function, and obtained the key to the problem in the analytical formula.

19. As shown in the figure, in the acute triangle ABC, points D and E are on the sides of AC and AB, AG⊥BC is at point G, AF⊥DE is at point F, EAF = ∠ GAC.

(1) Verification: △ ade ∽△ ABC;

(2) If AD=3 and AB=5, find the value of.

Analysis (1) shows that ∠ AFE = ∠ AGC = 90 Because of AG⊥BC and AF, it can be proved that ∠AED=∠ACB, and then △ ade ∽△ ABC;

(2)△ADE∽△ABC, and it is easy to prove △EAF∽△CAG, so it can be seen.

Solution: (1)∵AG⊥BC, AF⊥DE,

∴∠AFE=∠AGC=90，

∠∠EAF =∠GAC，

∴∠AED=∠ACB，

∫∠EAD =∠BAC，

∴△ADE∽△ABC，

(2) From (1), we can know: △ADE∽△ABC,

∴=

According to (1), ∠ AFE = ∠ AGC = 90,

∴∠EAF=∠GAC，

∴△EAF∽△CAG，

∴,

∴=

Comment on this question to test similar triangles's judgment. The key to solving the problem is to use similar triangles's judgment skillfully. This problem belongs to the medium type.

20. In all rectangles with equal areas, when one side of the rectangle is 1, the other side is 3.

(1) Let the lengths of two adjacent sides of a rectangle be x and y respectively.

① Find the functional expression of y about x;

② When y≥3, find the value range of x;

(2) Yuan Yuan said that the circumference of one rectangle was 6, and Fang Fang said that the circumference of one rectangle was 10. Do you think Yuan Yuan and Fang Fang are right? Why?

Analytic (1)① directly solve the rectangular area to get the relationship between y and x; ② directly use y≥3 to get the value range of x;

(2) Using the value of x+y and the discriminant of the root, we can get the answer directly.

Solution: (1)① From the meaning of the question: xy=3,

Then y =；;

② When y≥3, ≥3.

Solution: x ≤1;

(2) The circumference of a rectangle is 6,

∴x+y=3，

∴x+=3，

Finishing: x2-3x+3 = 0,

∵b2﹣4ac=9﹣ 12=﹣30，

The perimeter of a rectangle can be 10.

This topic mainly examines the application of inverse proportional function and the solution of quadratic equation in one variable. Correctly drawing the relationship between Y and X is the key to solving the problem.

2 1. As shown in the figure, in the square ABCD, the G point is on the diagonal BD (not coincident with the B point and the D point), the GE⊥DC is at the E point, and the GF⊥BC is at the F point, connecting the Ag.

(1) Write the quantitative relationship between the lengths of line segments AG, GE and GF, and explain the reasons;

(2) If the side length of the square ABCD is 1, ∠ AGF = 105, find the length of the straight line BG.

Analysis (1) Conclusion: Ag2 = Ge2+GF2. As long as it is proved that GA=GC and quadrilateral EGFC are rectangles, it can be deduced that GE=CF can be proved by Pythagorean theorem in Rt△GFC.

(2) Let BN⊥AG be n, and cut a point m on BN to make am = BM. Let an = X. AM=BM=2x and Mn = X. In Rt△ABN, according to AB2=AN2+BN2, we can get 1 = x2+(2x+x).

Solve x=, deduce BN=, and then solve the problem according to BG = bn ÷ cos 30;

Solution: (1) Conclusion: Ag2 = Ge2+GF2.

Reason: connect CG.

∵ quadrilateral ABCD is a square,

∴A and C are symmetrical about diagonal BD,

G point is on BD,

∴GA=GC，

∵GE⊥DC is at point E, GF⊥BC is at point F,

∴∠GEC=∠ECF=∠CFG=90，

∴ Quadrilateral EGFC is a rectangle,

∴CF=GE，

In Rt△GFC, ∫CG2 = GF2+CF2,

∴AG2=GF2+GE2.

(2) Let BN⊥AG be n, and cut a point m on BN to make am = BM. Let an = X.

∠∠AGF = 105，∠FBG=∠FGB=∠ABG=45，

∴∠AGB=60，∠GBN=30，∠ABM=∠MAB= 15，

∴∠AMN=30，

∴AM=BM=2x,MN=x，

At Rt△ABN, ∫AB2 = AN2+BN2,

∴ 1=x2+(2x+x)2，

The solution is x=,

∴BN=，

∴BG=BN÷cos30 =。

This topic examines the nature of the square, the judgment and nature of the rectangle, and the nature of the Pythagorean theorem, such as the 30-degree right triangle. The key to solving problems is to learn to add commonly used auxiliary lines and learn to use parameters to construct equations to solve problems, which belongs to the common questions in the senior high school entrance examination.

22. In the plane rectangular coordinate system, let the quadratic function y 1 = (x+a) (x-a- 1), where a ≠ 0.

(1) If the image of function y 1 passes through point (1, ﹣2), find the expression of function y 1;

(2) If the image of linear function y2=ax+b and the image of y 1 pass through the same point on the X axis, explore the relationship between real numbers A and B;

(3) It is known that points P(x0, m) and Q( 1, n) are on the image of function y 1. If m < n, find the value range of x0.

Analysis (1) According to the undetermined coefficient method, the resolution function can be obtained;

(2) According to the points on the function image that satisfy the resolution function, the answer can be obtained.

(3) According to the properties of quadratic function, we can get the answer.

Solution: The image of the (1) function y 1 passes through the point (1, ﹣2) and is obtained.

(a+ 1)(﹣a)=﹣2，

The solution is a =-2, a= 1,

The expression y=(x﹣2)(x+2﹣ 1) of the function y 1 is simplified to y = x2 ﹣ x ﹣ 2;

The expression y=(x+ 1)(x﹣2) of the function y 1 is simplified to y=x2﹣x﹣2.

To sum up, the expression of function y 1 is y = x2-x-2;

(2) When y=0, the solution of x2﹣x﹣2=0 is x1﹣1,x2=2,

The intersection of the image of y 1 and the X axis is (-1, 0) (2,0).

When y2=ax+b passes (-1, 0), -a+b = 0, that is, a = b；;

When y2=ax+b passes through (2,0), 2a+b=0, that is, b =-2a；;

(3) When P is on the left side of the symmetry axis, Y increases with the increase of X,

(1, n) and (0, n) are symmetrical about the symmetry axis,

From m < n, x0 < 0；;

When p is on the right side of the symmetry axis, y decreases with the increase of x,

Get 1 from m < n, x0 >.

To sum up: m < n, find the range of x0 < 0 or x0 > 1

This topic investigates the coordinate characteristics of points on quadratic function images. The key to solving (1) is to use the undetermined coefficient method. The key of solution (2) is to substitute the coordinates of points into the resolution function; The key of solution (3) is to make use of the properties of quadratic function, discuss it in categories and avoid omission.

23. As shown in the figure, it is known that △ABC is inscribed in ⊙O, point C is on the lower arc AB (not coincident with points A and B), point D is the midpoint of the chord BC, and the extension line of DE intersects with AC at point E, and the ray AO intersects with ray EB at point F, and intersects with ⊙O at point G, so ∞ is set.

(1) The following approximate data are obtained through drawing and measurement:

ɑ30 40 50 60

β 120 130 140 150

γ 150 140 130 120

Conjecture: the function expression of β about ι and the function expression of γ about ι, and give the proof:

(2) If γ = 135, CD=3, and the area of △ABE is four times that of △ABC, find the length of ⊙O radius.

Analysis (1) From the theorem of circumferential angle, we can get β = α+90, and then from the point where D is the midpoint of BC and DE⊥BC, we can get ∠ EDC = 90, and from the nature of the external angle of triangle, we can get ∠CED=α, so that we can know O, A, E and B.

(2) From (1) and γ = 135, we know that ∠ boa = 90, ∠ BCE = 45, ∠ BEC = 90, because the area of △ABE is four times that of △ABC, therefore, according to.

Solution: (1) conjecture: β = α+90, γ =-α+ 180.

Connect OB,

According to the theorem of the angle of circle, 2 ∠ BCA = 360 ∠ boa,

OB = OA，

∴∠OBA=∠OAB=α，

∴∠BOA= 180 ﹣2α，

∴2β=360 ﹣( 180 ﹣2α),

∴β=α+90 ,

D is the midpoint of de⊥bc BC province,

Oe is the perpendicular bisector of BC line,

∴BE=CE,∠BED=∠CED,∠EDC=90

∠∠BCA =∠EDC+∠CED，

∴β=90 +∠CED，

∴∠CED=α，

∴∠CED=∠OBA=α，

∴O, a, e, b four-point * * * cycle,

∴∠EBO+∠EAG= 180，

∴∠EBA+∠OBA+∠EAG= 180，

∴γ+α= 180 ;

(2) When γ = 135, the graph is as shown in the figure.

∴α=45 ,β= 135 ,

∴∠boa=90∠BCE = 45，

From (1), we can know that O, A, E and B are four * * * cycles.

∴∠BEC=90，

∵△ABE is four times the area of△ △ABC,

∴,

∴,

Let CE=3x, AC=x,

According to (1), BC=2CD=6,

∫∠BCE = 45，

∴CE=BE=3x，

According to Pythagorean Theorem, (3x)2+(3x)2=62,

x=，

∴BE=CE=3,AC=，

∴AE=AC+CE=4，

In Rt△ABE,

According to Pythagorean Theorem, AB2=(3)2+(4)2,

∴AB=5，

∫∠BAO = 45，

∴∠AOB=90，

In Rt△AOB, let the radius be r,

According to Pythagorean theorem, AB2=2r2,

∴r=5，

∴⊙ The length of the O radius is 5.

This topic examines the synthesis of circles, involving the knowledge of fillet theorem, pythagorean theorem, solving equations, the properties of the perpendicular line and so on. , with a high degree of comprehensiveness, which requires students to use what they have learned flexibly.