A few questions for 8th grade math, a few 8th grade math questions

Note: * denotes multiplying by sign;

1.From a2+b 2-6ab=0, we can know that a 2+b 2-2ab-4ab=0, i.e., (a-b) 2=4ab

In the same way, we can get a 2 + b 2 + 2 ab - 8ab = 0 i.e. (a + b ) 2 = 8 ab

Since a>b>0, (a+b) (b-a) can be replaced by above.

2.(x+y-z)(x-y-z)-(z-y-x)(x-y+z)=(x+y-z)(x-y-z)+(x+y-z)(x-y+z), which is an intentional common factor is: x+y-z

1 is derived from (a+3b)(a-2b)=0, a>b>0 to get that a+3b is not equal to 0, then a-2b=0, a=2b

a+b)/(b-a)=(2b+b)/(b-2b)=-3

2.(x+y-z)(x-y-z)=(z-x-y)(z-x+y), common factor (z-x-y).

3.Original formula = (-2) 2009*[1+(-2) 1]=(-2) 2009*(-1)=2 2009

4.From the known a=-1 2, the result of the equation is 17 4

5.After mentioning the common factor x, we get x(2x 2-x+m), where there is a common factor 2x+1, then (2x+1)(x+m)=2x 2-x+m, m=-2

6.The constant terms of the original formula are , -4, consider a group, a group of -2, -4 or a group of 1, -2, a group of 2, -4, the first allocation of the original formula is equal to y(xy+2)-2(xy+2)=(xy+2)(y-2); The second distribution is equal to xy(y-2)+2(y-2)=(xy+2)(y-2)

7.The original formula is equal to (a-4) 2-2(a-4) = (a-4) (a-4-2) = (a-4) (a-6) = a 2-10a + 24

8.The original formula is equal to (1+x)+x(1+x)+(1+x) 2+......x(1+x)^2009=(1+x)(1+x)+(1+x)^2+……x(1+x)^2009=(1+x)^2+x(1+x)^2+……x(1+x) 2009 continues to extract the common factor, and it is roughly seen that the square of the common factor is added to 1 every time it is extracted, and the final result is (1+x) 2010

Root number three 2 minus 5 4

3.Equilateral triangle.

1. Make the perpendicular bisector of the two points A and B, and the intersection point with L is.

2. The intersection point of the straight line BA and L is the principle that the difference between the two sides of the triangle is not greater than the third side.

3. It seems that you also know.

Proof: Angle AOB = Angle AOB + Angle FOB = 90° Angle FOB = Angle FOB + Angle BOE = 90°

So angular aof=angular boe

Angle FAO = Angle OBC = 45° (the angle between the diagonal line and the side length of the square) AO = BO

So AFO and BEO are congruent.

Wouldn't it be better after the doom was confirmed?

Congruent triangles correspond to equal sides.

oe=of

Because: AFO and BEO congruence (I'll write it if you want a process).

So: oe=of

Proof: The extension of the parallel line angle ac of the parallel line of the point e is a regular triangle at f abc.

The angle of ABC is ABC=60°, and the angle ACB=60°, then the angle of CEF ECF=angle FEC=60°

CEF is a regular triangle.

ef=ce, according to the title, apd and pef.

Angular apd = angular epf

ad=ef=ce

Angular DAP = Angular EFC = 60°

APD and FPE congruence.

So dp=pe

After D is df bc intersects ac to f, it is easy to know that adf is an equilateral triangle, then df=ce, so dfp ecp can be proved, so dp=pe

The area of this trapezoid is: [(upper bottom + lower bottom) * height ] 2=[(a+b)*(a+b)] 2=[(a+b) 2] 2;

The areas of the three small triangles are: a*b 2, a*b 2, c 2 2;

The trapezoidal area and the area of the three small triangles and the sum of the three small triangles, are: [(a+b) 2] 2=a*b 2+a*b 2+c 2 2;

Solution: a 2 + b 2 = c 2

That is, the Pythagorean theorem is verified.

The first question, you've already made it. In the second problem, when de is perpendicular to the x-axis, the shortest, then you can list the equation: t=20-3t solution, t=5, the line segment is the shortest.

In the third question, as long as the straight line found in the first two questions, as well as the straight line at the starting point, and the three lines are drawn, if the three lines are at the same point, then this point is the one sought.

Proof: The area of the three triangles is: 1 2AB, 1 2C and 1 2AB trapezoids have a top base, a bottom base, and a + B height

The area of the trapezoid is: 1 2 (a+b).

1 2(a+b) = 1 2ab + 1 2c +1 2ab, i.e. a +2ab + b = c +2ab

a²+b²=c²

The total area of the three triangles = 1 2ab + 1 2ab + 1 2c 2 = ab + 1 2c 2

Trapezoidal area = (a+b)(a+b) 2=1 2(a 2+2ab+b 2)=ab+1 2(a 2+b 2).

Because the two areas are equal, ab+1 2c 2=ab+1 2(a 2 + b 2).

i.e. a2+b2=c2

No, the average is greatly affected by individual values, and it is not possible to use the average to illustrate the size of the monthly salary.

Don't meet people like you who cross the river and tear down bridges again in your life, I think I'm unlucky when I meet you. It's best to do things next time.

1.The symmetry point g of e(f) is made through AC and connected to eg(fg).

It is calculated to be c (Pythagorean theorem).