Urgent! In the second year of junior high school, I took a math problem and asked for help!! Mathema

The original title is like this:

x-2y+z)(x+y-2z) (y-x)(z-x).

x+y-2z)(y+z-2x) (z-y)(x-y).

y+z-2x)(x-2y+z)fractions (x-z)(y-z)=?

Solution 1: Order: x-y=a, y-z=b, z-x=c, then a+b+c=0

So the original formula = [ac] [ab][-bc]+[ba] [bc][-ac]+[cb] [ac][-ab].

1/b*b+1/c*c+1/a*a

1/b^2+1/c^2+1/a^2

And because. 1/a+1/b+1/c

ab+bc+ca)/abc

x-y)(y-z)+(y-z)(z-x)+(z-x)(x-y)]/x-y)(y-z)(z-x）

And because. 1/a+1/b+1/c)^2

1/a^2+1/b^2+1/c^2+2/ab+2/ac+2/bc

1/a^2+1/b^2+1/c^2+2(a+b+c)/abc

1/a^2+1/b^2+1/c^2

So the original formula = 1 a 2 + 1 year old stupid b 2 + 1 c 2 = (1 hu with a + 1 b + 1 c) 2 = 1

Solution Ant Envy II:

Divide the original formula, and the three terms become the same denominator (x-2y+z)(x+y-2z)(y+z-2x).

Order: x-y=a, y-z=b, z-x=c, then a+b+c=0

Hence the first item of the molecule.

y-x)(z-x)(y+z-2x）

y-x)(z-x)(z-x)+(y-x)(z-x)(y-x)

c*c*(-a)+c*a*a

In the same way, the numerator of the second term = a*a*(-b) + a*b*b

In the same way, the numerator of the third term = b*b*(-c) + b*c*c

The denominator of the three items is the same, and the numerator sum is.

c*c*(-a)+c*a*a+a*a*(-b)+a*b*b+b*b*(-c)+b*c*c

a-b)(b-c)(c-a)

And the denominator = (x-2y+z)(x+y-2z)(y+z-2x) = (a-b)(b-c)(c-a).

So the original formula = (a-b)(b-c)(c-a) (a-b)(b-c)(c-a)=1

It is known that y=kx+b passes through the point a(-1,5), and the average mu is in a straight line y=-x, so the slope of y=kx+b and y=-x is equal, that is, k=-1 substitutes, and substituting the point a(-1,5) into y=-x+b.

5=1+bb=4

So the analytic expression of the straight line is y=-x+4

Substituting the point b(m,-5) into y=-x+4 to get Xun Xiaosen.

5=-m+4

m=9 Let the line intersect the x-axis at the point c(4,0).

Then AOB area = AOC area + BOC area.

AOC face caution or product = |4*5|/2=10

BOC area = |4*(-5)|/2=10

So AOB area = 10 + 10 = 20

Find it by area equality.

Let the distance be m, then:

12*5/2=（12+5+13）*m/2

The distance is 2

The center of a triangle inscribed circle.

Using the area, connect DA, DB, and DC into three triangles, and the total area is the entire right triangle.

First of all, I have to say that you didn't learn the triangle congruence problem well;

Second, there are some errors in your question input, which means that you are not serious enough, which also leads to the fact that I can only answer 3 of your remaining 4 questions, because I can't guess that you ** made a mistake in question 8.

Here is the answer to the third question. For your convenience, it was made into the form of **.

The sum of the outer angles of 1 is 360° and should be 9 sides.

2 rectangles are also parallelograms, so the area is 4*5=20

3,38 degrees.

Question 4: Because the angle ebo+beo=fce+beo=90, the angle ebo=ecf and ob=oc

So the triangle beo congruent triangle ocg

So the two sides are equal.

The sum of the outer angles of 1 is 360° and should be 9 sides.

2. The rectangle is also a parallelogram, so the area is 4*5=203,38 degrees.

4. Because the angle ebo+beo=fce+beo=90, the angle ebo=ecf and ob=oc

So the triangle beo congruent triangle ocg

So the two sides are equal.

(1) Because of folding, af=ad=10

There is the Pythagorean theorem, which gives bf=6

Because because of folding, ef=de

Let ef=x, then ec=cd-de=8-x

And because cf=4, there is the Pythagorean theorem, get.

8-x) + 4 = x.

x=5ce=3

2）s =4*3*

x-2y-3+(2x-3y-5) 2=0 gives x-2y-3=0

2x-3y-5=0

x=1,y=-1

x-8y=1+8=9

So the square root of x-8y is plus or minus 3, and the cube root is 9 under the 3rd root number

Because the perfect square and the root sign are not negative.

So x-2y-3=0(1).

2x-3y-5=0（2）

x-y-2=0

x=y+2（3）

3) Substitution (1).

y+2-2y-3=0

y=-1x=-1+2=1

x-8y=1+8=9

The square root of x-8y is 3, and the cubic root is 9 under the 3rd root number

Because x-2y-3+(2x-3y-5) 2=0, x-2y-3=0 a.

2x-3y-5=0 b.

A 2 = 2x-4y-6 = 0 C formula.

c-b=-y-1=0

y=-1 is substituted for a.

x+2=3 so x=1

x-8y=9

x-8y=3

3^√x-8y=3^√9

Let it take 2x hours for A to be filled alone, then it will take 3x hours for B to be filled separately by the inscription of 2 2x+3 3x=1

The solution is x=23x=6

The speed ratio is 3:2 and set to 3t and 2t

Time = ((3+2)*2+2) 2=6h

Let the time of pipe A be x, and the time of pipe B is y

2(1/x)+y=1

2:3=x:y (can be changed to 2y=3x) to solve by itself.