Comprehensive training at the end of the ninth grade of Beijing Normal University I. Fill in the blanks:

1, the equation (x+ 1)(x+2)=3 is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

2. The image of the inverse proportional function passes through the point (–2,4), so the analytical formula of this inverse proportional function is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

3. The two sides of an isosceles triangle are 4cm and 9cm respectively, and the circumference is _ _ _ _ _ _ cm.

4. If the outer angle of the isosceles triangle is 80, then the base angle = _ _ _ _ _.

5. The quadratic equation of the root is _ _ _ _ _ _ _ _.

6. The unary quadratic equation 2x(kx-4)-x2+6=0 has a real root, and the largest integer of k is _ _ _ _ _.

7. Given that one root of the quadratic equation of X is–2, then k = _ _ _ _ _ _ _ _

Second, choose:

1. It is known that y is inversely proportional to x=4, and y=, then the functional relationship between y and x is ().

a、y= B、y= C、y= D、y=

2, known to be completely flat, then the value of k is ()

A, 2 B, 2 or -2 C, -2 D, 1 or-1

3. The figure with both axial symmetry and central symmetry is _ _ _ _ _ _ _ _.

A, regular triangle b, rectangle c, parallelogram d, isosceles trapezoid

4. If xy=b and (x+y)2 is equal to ().

a 、( a+b)2 B、a2+b2 C、b(ab+2) D、ab(b+2)

5, in the trapezoid ABCD, AD//BC, ∠ A = 90, BD ⊥ CD, AB = AD, then ∠C is _ _ _ _ _. A, B, 45 degrees Celsius, 60 days, 22: 05

6. In trapezoidal ABCD, where AD//BC, AD=m, BC=n, E and F are the midpoint of diagonal AC and BD respectively, then EF = _ _ _ _ _ _ _ a、B、n-m C、D、

7. As shown in the figure, O is the intersection of diagonal AC and BD of parallelogram ABCD, and EF passes through point O and intersects with sides AD and BC at points E and F respectively. If FB=DE, there are () A, 2 pairs of B, 3 pairs of C, 5 pairs of D and 6 pairs in congruent triangles.

8. If the algebraic expressions 4x2-2x-5 and 2x2+ 1 are reciprocal, the value of x is ().

A, 1 or-1.5 B, 1 or -C,-1 or d, 1 or 1.5.

9, for the height of the triangle ABC AD, angle bisector AE, midline AF, among these three may be outside the triangle is ().

It's possible.

10. If the image of the function passes through the point (-1 2), then the image of the function must be at ().

(a) first and second quadrants (b) third and fourth quadrants

(c) first and third quadrants (d) second and fourth quadrants

Third, solve the following equations with appropriate methods.

Fourth, answer questions.

1. An opaque box contains eight red balls and two yellow balls. Except for the different colors, each ball is the same. Take any ball out of the box and put it back in the box.

(1) Every time you touch the ball, do you think it is more likely to touch the red ball or the yellow ball?

(2) Xiaoming touched it five times according to the above method, and all he touched were yellow balls. Is this possible? Why?

According to the above facts, please talk about the relationship between the probability and frequency of random events.

2. The railway distance between A and B is 65,438+600 km. After the technical transformation, the train speed is increased by 20 km/h compared with that before the speed increase, and the travel time from place A to place B is shortened by 4 hours. Under the existing conditions, the safe running speed of this railway should not exceed1400 km/h. Please use your mathematical knowledge to explain whether the train can still run under the existing conditions of this railway.

3. As shown in the figure, it is known that in △ABC, AD=AE, BD⊥AC is in D, CE⊥AB is in E for verification, ∠DBC=∠ECB.

4. As shown in figure △ABC, D is the midpoint of df⊥ac de⊥ab BC, and the vertical foot is E, F, be = CF Verification: DE=DF. Answer:

fill (up) a vacancy

1、x2+3x- 1=0， 1，- 1

2、

3. The waist length of isosceles triangle is 9cm, the bottom is 4cm and the circumference is 22cm.

4. Because one outer angle of an isosceles triangle is 80, the adjacent inner angle must be 100, and only the vertex angle. Therefore, there is only one case in this question, and both the inner angle and the derivable base angle are 40.

5、

6、 1

7，k=4。 Substitute x=-2 into the original equation, construct a quadratic equation about k, and then solve it.

Second, choose

1、A

∵ y is inversely proportional to and will be used as an independent variable. Let the analytical formula be y=,

When x=4, y=,

∴ ,∴

Choose a.

2.B

3.B

4.c∶∴y2+x2 = a(xy)2，

∴(x+y)2-2xy=a(xy)2(x+y)2=a(xy)2+2xy=ab2+2b=b(ab+2)

5. According to the condition ∠ Abd = ∠ ADB = 45,

∴ ∠DBC=45，

And BD⊥DC,

∴ ∠C=45，

So choose B.

6. (Assuming BC & gtAD)

Connect DE and extend BC to point h,

Δ AED Δ CEH (ASA) can be easily obtained.

∴ DE=EH，AD=CH=m，

And ∵ DF=FB,∴ EF is the center line of δδDBH,

∴ EF= BH

= (BC-CH)

= (n-m)

If AD> is in BC, EF= (m-n)

So choose d,

7、D 8、B 9、A

10、D

The function passes through points (-1, 2), ∴, k+ 1 =-2.

Third, solve the equation

(1) solution: 4(x+2)2-9(x-3)2=0.

[2(x+2)+3(x-3)][2(x+2)-3(x-3)]= 0

(5x-5)(-x+ 13)=0

5x-5=0 or -x+ 13=0.

∴x 1= 1，x2= 13

(2) Solution: x2-2x=-

(first become a general form)

So, that is to say,

Fourth, answer questions.

1, (1) is very likely to touch the red ball, (2) it is possible, because this is an uncertain event, and the result is unpredictable every time it touches the ball. (3) When there are many experiments, the frequency of an event will be stable near the corresponding probability. So we can use the frequency of an event to estimate the probability of this event through many experiments.

2. Solution: Let the speed of the train before the speed increase be x km/h, then the speed after the speed increase is (x+20) km/h,

According to the meaning of the question:

Arrangement: x2+20x-8000=0, solve this equation: x 1=- 100, x2=80,

It is verified that X 1 =- 100 and X2 = 80 are the roots of the original equation.

But x=- 100 is irrelevant, so it is omitted, ∴ x=80.

At this time, 80+20= 100, that is, the speed of the train after speeding up is 100 km/h,

Because the safe running speed of this railway should not exceed 140 km/h,

100 & lt； 140, so the train can speed up again under the existing conditions of this railway.

3. Proof: ∵BD⊥AC, CE⊥AB (known)

∴∠BDA=900, ∠CEA=900 (vertical definition)

∴∠BDA=∠CEA (equivalent substitution)

In △ABD and △ACE.

∵

∴△ABD≌△ACE(ASA)

∴BD=EC (the corresponding sides of congruent triangles are equal)

In Rt△BCE and Rt△CBD.

∵

Rt△BCE≌Rt△CBD(HL)

∴∠DBC=∠ECB (the corresponding angles of congruent triangles are equal).

4. Proof:

∵DE⊥AB is in E, DF⊥AC is in F.

∴∠BED=∠CFD=90

In Rt△BED and Rt△CFD.

BE = CF，BD=CD

∴Rt△BED≌Rt△CFD

∴DE=DF