\(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2+y^2-2xy\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay \(x-y=7\)vào biểu thức, ta có:

\(A=7^2+2.7+37\)

\(=49+14+37\)

\(=100\)

A= X(X-2)+Y(Y-2) -2XY+37

= X²-2X+Y²-2Y -2XY +37

=(X²-2XY+Y²)+(2X-2Y)+37

=(X-Y)²+2(X-Y)+37

Thế số vào là xong :) :)

So sorry ...... e ko giúp chị được vì ..... e mới lên lớp 6 <3 

Mọi người k ủng hộ e được ko ạ !!! Nếu được e cảm ơn vì đã động viên e nha ###

Ai đi qua cho em xin 1 k để chuẩn bị cho kì thi tuyển sinhhhh ạ !!!!

1) \(x^2-2x-4y^2-4y\)

\(=\left[x^2-\left(2y\right)^2\right]-\left(2x+4y\right)\)

\(=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

2) \(x^4+2x^3-4x-4\)

\(=\left(x^4-4\right)+\left(2x^3-4x\right)\)

\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)

\(=\left(x^2-2\right)\left(x^2+2+2x\right)\)

3) \(x^2\left(1-x^2\right)-4x+4x^2\)

\(=x^2\left(1+x\right)\left(1-x\right)+4x\left(x-1\right)\)

\(=x^2\left(1+x\right)\left(1-x\right)-4x\left(1-x\right)\)

\(=\left(1-x\right)\left[x^2\left(1+x\right)-4x\right]\)

1/ \(3x^2+6x+3-3y^2=3x^2+3x+3x+3-3y^2\)

\(=3\left(x^2+2x+1-y^2\right)\)

\(=3\left[\left(x^2+2x+1\right)-y^2\right]\)

\(=3\left[\left(x+1\right)^2-y^2\right]\)

\(=3\left(x+1-y\right)\left(x+1+y\right)\)

2/ \(25-x^2-y^2+2xy=5^2-\left(x^2+y^2-2xy\right)\)

\(=5^2-\left(x-y\right)^2\)

\(=\left[5-\left(x-y\right)\right]\left(5+x+y\right)\)

\(=\left(5-x+y\right)\left(5+x+y\right)\)

3/ \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)

\(=3\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left[3-\left(x-y\right)\right]\)

\(=\left(x-y\right)\left(3-x+y\right)\)

a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)\)

\(=\left(x-3\right)\left(x^2-2^2\right)\)

\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)

b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)

\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)

\(=\left(x-2\right)\left(x+2+x-2\right)\)

\(=\left(x-2\right)2x\)

c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

a) => x².(x-3)-4(x-3)=(x-3)(x²-4)=(x-3)(x-2)(x+2)

b) => (x+2)(x-2)+(x-2)²=(x-2)(x+2+x-2)=2x(x-2)

c) => x³+27-(4x²+12x)=(x+3)(x²-3x+3)-4x(x+3)=(x+3)(x²-3x+3-4x)=(x-3)(x²-7x+3)

nó dễ ợt mà -_- 

x²y²+4xy+4=(xy+2)² xong :)) 

a)\(x^4-2x^3+2x-1=x^4-x^3-x^3+x+x-1\)

\(=x^3\left(x-1\right)-x\left(x^2-1\right)+\left(x-1\right)\)

\(=x^3\left(x-1\right)-x\left(x-1\right)\left(x+1\right)+\left(x-1\right)\)

\(=x^3\left(x-1\right)-\left(x^2+x\right)\left(x-1\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left[x^3-\left(x^2+x\right)+1\right]\)

\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)

\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)

\(=\left(x-1\right)\left(x-1\right)\left(x^2-1\right)\)

\(=\left(x-1\right)\left(x-1\right)\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)^3\left(x+1\right)\)

b)\(x^4+2x^3+2x^2+2x+1=x^4+x^3+x^3+x^2+x^2+x+x+1\)

\(=x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+x^2+x+1\right)\)

\(=\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)

\(=\left(x+1\right)\left(x+1\right)\left(x^2+1\right)\)

\(=\left(x+1\right)^2\left(x^2+1\right)\)

a) =

\(x^4-x^3-x^3+x^2-x^2+x+x-1=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(x^3-x^2-x+1\right)\)

a) (x+y)²-y²=x²+2xy+y²-y²=x²+2xy=x(x+2y)  (đpcm)

2 a )

4a²+4a+2=(2a)²+2.2a+1+1=(2a+1)²+1

vì (2a+1)² lớn hơn hoặc = 0 với mọi a nên (2a+1)²+1 lớn hơn hoặc = 1 

dấu ''='' xảy ra khi 2a+1=0<=>a=-1/2