1、(8-9).(9- 10).( 10- 1 1)......(2004-2005)=_______

2. (7/9+7/18-5/6) * 36-1.45 * 6+3.95 * 6 = (process)

3、- 1/2+ 1/3=____

4. The fourth power of-5+the fourth power of 5+(-11999) and the fifth power of *( 1999)) = _ _

5. The 2008th power of (-2) * the 2009th power of (-0.5) = _ _

1. What is the value of (1) given as x*x-5x+ 1=0? x*x+ 1/(x*x)？ (2).？ What is the value of the reciprocal of the fourth power of X plus the fourth power of X?

2.？ Given (2x+1) (3x-2) (ax * x+bx+c) = 6x * x * x+kx * x+LX * x-3x-2, how to find k.l and a.b.c?

3. Given X-Y ≠ 0. X * X-X = 7 and Y * Y-Y = 7. Find the value of X * X * X+Y * Y+X * XY+XY * Y?

4. Given that x*x+y*y-2x+4y+5=0, what is the value of 4/(y+ 1/x)?

5. Given a*a*a-3ab-4b*b=0(a≠0), what is the value of (2a+b)/(a-b)?

6. if x.y.z satisfies (y+z)/x=(z+x)/y=(x+y)/z=k, find the value of k.

1. It is known that one root of the unary quadratic equation 2mx 2-2x-3m-2 = 0 is greater than 2, and the other root is less than 1, so the value range of the number m is realistic.

2. When k is what value, the two absolute values of the unary quadratic equation 2 (k+3) x 2+4kx+3k-6 = 0 are equal.

3. The unary quadratic equation MX 2-2mx+m-5 = 0 has a positive root and a negative root, and the resignation range of the number m is sought.

4. Find the value or range of k when the quadratic function y = x 2-(12-k) x+12 meets the following conditions respectively.

1)y has a completely flat representation.

2) The axis of symmetry is to the right of point (3,-1).

3) What is the minimum value?

4) The vertex position is the highest.

5) The area of a triangle with three intersections with X and Y axes as vertices is 6k.

5. if x/(a-b)=y/(b-c)=z/(c-a), then x+y+z=

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

Answer:

1、- 1^ 1997=- 1

2、(7/9+7/ 18-5/6)*36- 1.45*6+3.95*6

=[(7*2+7- 15)/ 18]*36-6*( 1.45-3.95)

=6? / 18*36-27/5*6-6*(-2.5)

= 12-(- 15)

= 12+ 15=27

3、- 1/2+ 1/3=-3/6+2/6=- 1/6

4、-5^4+(-5)^4+(- 1/ 1999)^5*(- 1999)^5

=-(5^4)+(5^4)+ 1/(- 1999^5)*(- 1999^5)

=0+ 1= 1

The 2008th power of 5 (-2) and the 2009th power of *(0.5).

=(-2)^2008*(- 1/2)^2009

=(2^2008)*(- 1/2)^2008*(- 1/2)

=(2^2008)* 1/(2^2008)*(- 1/2)

= 1*(- 1/2)

=-0.5

1. divided by two known sides at the same time, x+ 1/x=5, divided by the square of two sides, x 2+2+1/x 2 = 25,

( 1)x^2+ 1/x^2=23.

(2)(x^2+ 1/x^2)^2=529,x^4+2+ 1/x^4=529，

x^4+ 1/x^4=527

2.left =(6x 2-x-2)(ax2+bx+c)= 6ax 4+(6 b-a)x3+(6c-2a-b)x2-(2 b+c)x-2c，

The comparison coefficients on the right are: a = 1, k = 6b-a, l = 6c-2a-b, 2b+c =-3, c = 1,

a= 1，b=-2，c= 1，k=- 13，l=6

1, solution: Let y = 2mx 2-2x-3m-2, then the two roots α and β of the equation y=0 are the abscissas of the two intersections of the parabola y = 2mx 2-2x-3m-2 and the X axis.

When m>0, α < 1, β > 2, the necessary and sufficient conditions are: f (1).

When m<0, α < 1, β > 2, the necessary and sufficient conditions are: f( 1)>0, f(2)>0, that is, m.

Therefore, the range of m is: m.

2. Let the two roots of the equation be X 1 and X2.

When X 1+X2=0 and -4K/2(K+3)=0, K = 0；;

When X 1=X2, (4k) 2-8 (k+3) (3k-6) = 0, k = 3 or K=-6.

So when k = 0, k = 3 and k =-6, the absolute values of the two roots of the equation are equal. ?

3. Let the two roots of the equation be X 1 and X2.

△& gt； 0, that is, 4m 2-4m (m-5) > 0, so m>0

x 1 * X2 & lt； 0, namely (m-5)/m.

So, 0

4、

If 1)y is completely flat, then y = (x-√12) 2 = x2-(2 √12) x+12.

Therefore, 2√ 12= 12-k,

k= 12-4√3

2) Is the axis of symmetry on the right side of the point (3,-1)?

Namely: (12-k)/2 >; three

So k> six

3) What is the minimum value?

Namely: [48-( 12-k) 2]/4 = 3.

Therefore, k=6 or k= 18.

4) The ordinate of the vertex is [48-( 12-k) 2]/4,

Namely:-1/4 * k 2+6k-24. When k=6/( 1/2)= 12,-1/4 * k 2+6k-24 has the maximum value.

So, when k= 12, the vertex position is the highest? .

5) When x=0, y= 12,

What is the area of the triangle with the X axis and the Y axis as the vertices?

Therefore, | x 1-x2 | = k

X 1+X2= 12-k，X 1*X2= 12，

Therefore, k 2 = (12-k) 2-4 * 12.

So, k=4.

5. let x/(a-b) = y/(b-c) = z/(c-a) = k.

So x=(a-b)*k, y=(b-c)*k, and z = (c-a) * k.

Then x+y+z = (a-b) * k+(b-c) * k+(c-a) * k.

Therefore, x+y+z=0.

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

Here are some painting questions:

1, as shown in figure 1, OA = OB, OC = OD, ∠ O = 50, ∠ D = 35, then ∠AEC is equal to ().

2. As shown in Figure 2, AB=AC is known. To get △ Abe △ ACD from "ASA", one condition should be added: ().

3. As shown in Figure 3, if AD=CB, ∠A=∠C, then △ AOD △ COB can be interpreted according to ().

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

The process of writing the following questions

4. As shown in Figure 4, it is a big rectangle composed of 10 identical small rectangular pieces of paper. After measurement, the width of this big rectangle is 50 cm. Try to find the area of a small rectangle.

5. As shown in Figure 4, if 4, AC and BD intersect at point O, OA=OC, ∠A=∠C, then OB=OD?

6. As shown in Figure 6, in a square ABCD, AC is diagonal, and E is a point above AC, connecting EB and ED.

(1) Description: △ bec △ dec

(2) extend BE-to-AD to f, and find the number of times when ∠ bed = 120 ∠EFD.

7. As shown in Figure 7, AF bisects ∠BAC, BC⊥AF, the vertical foot is E, point D and point A are symmetrical about point E, and PB is compared to P and M by line segments CF and AF respectively.

(1) means AB=CD.

(2) If ∠BAC=2∠MPC, please judge the quantitative relationship between ∠F and ∠MCD and explain the reasons.

8. Two groups of quadrilaterals with equal adjacent sides are called orthomorphism. As shown in Figure 8, in orthomorphic ABCD, AB=AD, BC=DC, AC and BD intersect at O point.

(1) Description: ①△ABC?△ADC；; ; ②OB=OD,AC⊥BD

(2) If AC=6 and BD=4, find the area of conformal ABCD.

9. As shown in Figure 9, in the right-angled trapezoidal ABCD, ∠ ABC = 90, AD//BC, AB=BC, E is the midpoint of AB, and CE⊥BD.

(1) Description

(2) Description: AC⊥DE

(3) Is △ DBC an isosceles triangle? And explain why.

10, as shown in figure 10, where d is a point on the AB side of △ABC, e is the midpoint of AC, and FC//AB.

(1) Description △ ade △ CFE

(2) If AB=9 and FC=7, find the length of BD.

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

This is the answer:

(1) proves that ∵ quadrilateral ABCD is a square.

∴BC=CD,∠ECB=∠ECD=45

EC = EC again?

∴△ABE≌△ADE

(2)∫△ABE?△ADE

∴∠BEC=∠DEC= 1/2∠BED

∠BED= 120 ∴∠BEC=60 =∠AEF

∴∠EFD=60 +45 = 105？

7. Solution: (1) Proof: ∫AF bisection ∠BAC,? .

∴∠CAD=∠DAB= 1/2∠BAC.

∫d and a are symmetric about e, and∴ E is the midpoint of AD.

∴bc ∵bc⊥ad is the middle vertical line of AD, ∴ AC = CD.

In Rt△ACE and Rt△ABE, note that AC=CD can also be obtained by adjoint congruence.

∠CAD+∠ACE=∠DAB+∠ABE=90，？ ∠CAD=∠DAB。

∴∴∠ace=∠abe。 Note: AC = AB can also be obtained if it is congruent.

∴AB=CD.？

(2)∫∠BAC = 2∠MPC，？ ∠∠BAC = 2∠CAD，∴∠ MPC = ∠ CAD。

∵AC=CD,∴∠CAD=∠CDA,∴∠MPC=∠CDA.？

∴∠MPF=∠CDM.？

* AC = ab，AE⊥BC，∴ CE=BE。 Note: CE = BE can also be obtained by congruence.

∴AM is the middle vertical line of BC, ∴ CM=BM. Note: CM = BM can also be obtained by syndrome congruence.

∵EM⊥BC, ∴EM bisects ∠CMB, (isosceles triangle with three lines in one)?

∴∠CME=∠BME .. Note: ∞∠CME =∠ Can I get BME with the same certificate?

∠∠BME =∠PMF，

∴∠PMF=∠CME，

∴∠MCD=∠F (sum of angles in a triangle). Note: ∠MCD=∠F can also be obtained by proving triangle similarity.

8. Prove: (1)① in △ABC, △ADC,

AB=AD，BC=DC，AC=AC，

So ABC?△ADC；; ; ?

② Because of △ ABC△ ADC,

So horn abalone = = horn knife,

In the delta package, delta track,

AB=AD, angle BAO= = angle DAO, ao = ao,

So △ Bao△ Dao,

So ob = od?

Because angle BAO= = angle DAO,

So AC is the bisector of the angle BAC,

Because OB=OD,

So AC⊥BD (the distance from the bisector of the angle to both sides of the angle is equal)

(2) Because of △ ABC△ ADC,

So the area of Zheng-shaped △ABCD is twice that of△ ABC.

And because △ ABC area s = AC * BD/2 = 6 * 4/2 =12,

The area of Zheng ABCD is 2s = 24.

9. Proof: (1) ∠ ABC = 90, BD⊥EC.

∴∠ 1 and ∠3 are complementary, ∠2 and ∠3 are complementary,

∴∠ 1=∠2

∠∠ABC =∠DAB = 90 °, AB = AC

∴△BAD≌△CBE

∴AD=BE…

(2)E is the midpoint of AB,

∴EB=EA

From ( 1)AD=BE: AE=AD。

∫ AD ∨ BC

∴∠7=∠ACB=45

∵∠6=45

∴∠6=∠7

From the properties of isosceles triangle, we can get: EM=MD, AM⊥DE.

That is, AC is the middle vertical line of line segment ED.

(3)△DBC is an isosceles triangle (CD=BD)

The reason for this is the following：

From (2): CD=CE

From ( 1): CE=BD

∴CD=BD

△ DBC is an isosceles triangle.

The picture is a little unclear. Open the big picture yourself and copy it to the computer. . I don't know if you can understand some questions, but they should all be relatively basic. Some things you can only learn when you are in junior high school. Anyway, take your time and get to know them carefully. )