a.

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

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

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

= 2x^2 + 2y^2

b.

(x^2 + y^2 - 2xy) - (x^2 + y^2 + 2xy)

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

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

= -4xy

`x/(x+y) + (2xy)/(x^2-y^2) - y(x+y)`

`= (x(x-y))/(x^2-y^2) + (2xy)/(x^2-y^2) - (y(x-y))/(x^2-y^2)`

`= (x^2 - xy + 2xy - xy + y^2)/(x^2-y^2)`

`= (x^2+y^2)/(x^2-y^2)`

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

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

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

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

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

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

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

`@` `\text {Ans}`

`\downarrow`

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

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

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

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

giúp mk với nhé

sáng mai nộp rồi 

ai nhanh tay mk sẽ k cho

[(x²-2xy+2xy²).(x+2y)-(x²+4y²).(x-y)]2xy

=( x³ + 2x²y-2x²y-4xy²+2x²y²+4xy³-x³+x²y-4xy²+4y³ )2xy

=2xy(2x²y²-8xy²+4xy³+x²y+4y³)

= 4x³y³-16x²y³+8x²y⁴+2x³y²+8xy⁴

Trả lời:

[ ( x² - 2xy + 2xy² ) ( x + 2y ) - ( x² + 4y² ) ( x - y ) ] 2xy

= [ ( x³ + 2x²y - 2x²y - 4xy² + 2x²y² + 4xy³ ) - ( x³ - x²y + 4xy² - 4y³ ) ] 2xy

= ( x³ + 2x²y - 2x²y - 4xy² + 2x²y² + 4xy³ - x³ + x²y - 4xy² + 4y³ ) 2xy

= ( x²y - 8xy² + 2x²y² + 4xy³ + 4y³ ) 2xy

= 2x³y² - 16x²y³ + 4x³y³ + 8x²y⁴ + 8xy⁴

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

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

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

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

\(=\dfrac{y}{x-y}-\dfrac{x}{x-y}\)

\(=\dfrac{y-x}{x-y}=\dfrac{-\left(x-y\right)}{x-y}=-1\)

a: \(=x-\dfrac{3}{2}+2y\)

b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)