Bài 1:

a, \(\left(x-y+2z\right)^2=x^2+y^2+4z^2-2xy-4yz+4zx\)

b, \(\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)=\left(4x^2-9\right)\left(4x^2+9\right)=16x^4-81\)

a)\(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}=\dfrac{\left(x-y\right)^2\left[3\left(x-y\right)^2+2\left(x-y\right)-5\right]}{\left(x-y\right)^2}=3x^2-6xy+3y^2+2x-2y-5\)

b) \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}=\dfrac{\left(x-2y\right)^3}{\left(x-2y\right)^2}=x-2y\)

c) \(\dfrac{x^3+y^3}{x+y}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}=x^2-xy+y^2\)

a: \(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}\)

\(=\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(x-y\right)^2}\)

\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)

b: \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}\)

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

=x-2y

c: \(\dfrac{x^3+y^3}{x+y}\)

\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}\)

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

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

\(=\left(2x+1\right)^2+2\cdot\left(2x+1\right)\cdot1+1^2\)

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

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

b) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)\)

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

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

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

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

\(=4y^2\)

a/ \(\left(-2x^2y\right)5xy^4\)

\(=-10x^3y^5\)

tự lm tiếp, rất dễ

a) Ta có: \(\left(-2x^2y\right)\cdot\left(5xy^4\right)\)

\(=\left(-2\cdot5\right)\cdot\left(x^2\cdot x\right)\cdot\left(y\cdot y^4\right)\)

\(=-10x^3y^5\)

b) Ta có: \(\left(\dfrac{27}{10}x^4y^2\right)\cdot\left(\dfrac{5}{9}xy\right)\)

\(=\left(\dfrac{27}{10}\cdot\dfrac{5}{9}\right)\cdot\left(x^4\cdot x\right)\cdot\left(y^2\cdot y\right)\)

\(=\dfrac{3}{2}x^5y^3\)

c) Ta có: \(\left(\dfrac{1}{3}x^3y\right)\cdot\left(-xy\right)^2\)

\(=\dfrac{1}{3}x^3y\cdot x^2y^2\)

\(=\dfrac{1}{3}x^5y^3\)

a: \(=x^2-xy+xy+y^2=x^2+y^2=100\)

b \(=x^3-xy-x^3-x^2y+x^2y-xy=-2xy=-2\cdot\dfrac{1}{2}\cdot\left(-100\right)=-1\cdot\left(-100\right)=100\)

a)` x(x - y) + y(x + y) `

`=x^2-xy+xy+y^2`

`=x^2+y^2`(1)

thay x= -6 ; y= 8 vào 1 ta đc

\(\left(-6\right)^2+8^2=36+64=100\)

b)`) x(x^2 - y) - x^2 (x + y) + y (x^2 - x) `

`=x^3-xy-x^3-xy+yx^2-xy`

`=\(-3xy+yx^2\)(2)

thay `x= 1/2 và y = -100` ta đc

\(-\dfrac{3.1}{2}.\left(-100\right)+\dfrac{\left(-100\right).1}{2}=150-50=100\)

=100

a: \(x\left(x-y\right)+y\left(x+y\right)\)

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

\(=x^2+y^2\)

=100

b: \(x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)\)

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

\(=-2xy\)

Viết sai đề 😡

a) A = 2x^2 + 2y^2

a, \(A=\left(x-y\right)^2+\left(x+y\right)^2\)

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

\(=2x^2+2y^2\)

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

b) \(B=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\\ =4x^2-4x+1-2\left(4x^2-12x+9\right)+4\\ =4x^2-4x+1-8x^2+24x-18+4\)

    \(=-4x^2+20x-13\)

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

`= x^2 - 2xy - y^3 + 2xy`

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)

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

\(=9x^2-6x+1-2x^2+x-6x+3\)

\(=7x^2-11x+4\)