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)`

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

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

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

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

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

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:

$(x-1)(x^2+x+1)-x^3-6x=11$

Dùng $(x-1)(x^2+x+1)=x^3-1$:

$x^3-1-x^3-6x=11$

$-6x-1=11$

$-6x=12$

$x=-2$

Vậy $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

Điều kiện: $x\ne-2,-1$.

$(x-1)(x+1)=(x+2)^2$

$x^2-1=x^2+4x+4$

$-4x=5$

$x=-\dfrac54$

Vậy $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

Điều kiện: $x\ne-2,-1$.

$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$

$x+1+2x+4=0$

$3x+5=0$

$x=-\dfrac53$

Vậy $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

Điều kiện: $x\ne0,3$.

$\dfrac{9-x^2}{x(x-3)}=1$

$9-x^2=x(x-3)$

$9-x^2=x^2-3x$

$2x^2-3x-9=0$

$(2x+3)(x-3)=0$

$x=-\dfrac32$ hoặc $x=3$

Nhưng $x=3$ không thỏa điều kiện.

Vậy $x=-\dfrac32$.

Ta có: \(\frac{3}{x^2+2xy+y^2}+\frac{4}{2xy-x^2-y^2}+\frac{5}{x^2-y^2}\)

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

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

\(=\frac{3x^2-6xy+3y^2-4x^2-8xy-4y^2+5x^2-5y^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

\(=\frac{4x^2-14xy-6y^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

Ta có: \(\frac{3}{x^2+2xy+y^2}+\frac{4}{2xy-x^2-y^2}+\frac{5}{x^2-y^2}\)

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

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

\(=\frac{3x^2-6xy+3y^2-4x^2-8xy-4y^2+5x^2-5y^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

\(=\frac{4x^2-14xy-6y^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

`@` `\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: \(\left(2x^2y-3xy+4xy^2\right):2xy\)

\(=\frac{2x^2y}{2xy}-\frac{3xy}{2xy}+\frac{4xy^2}{2xy}\)

\(=x-\frac32+2y\)

b: \(\frac{1}{xy}-\frac{x^2-1}{y^2-xy}\)

\(=\frac{1}{xy}-\frac{x^2-1}{y\left(y-x\right)}\)

\(=\frac{y-x}{xy\left(y-x\right)}-\frac{x\left(x^2-1\right)}{xy\left(y-x\right)}=\frac{y-x-x^3+x}{xy\left(y-x\right)}=\frac{-x^3+y}{xy\left(y-x\right)}\)

c: \(\left\lbrack\frac{x}{xy-y^2}-\frac{2x-y}{x^2-xy}\right\rbrack:\left(\frac{1}{x}-\frac{1}{y}\right)\)

\(=\left\lbrack\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x\left(x-y\right)}\right\rbrack:\frac{y-x}{xy}\)

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

=-1