Ta có:

 VT=(x2+y2)2−(2xy)2VT=(x2+y2)2−(2xy)2

=(x2+y2−2xy)(x2+y2+2xy)=(x2+y2−2xy)(x2+y2+2xy)

=(x−y)2(x+y)2=VP=(x−y)2(x+y)2=VP

⇒đpcm⇒đpcm

Bexiu bị j ấy

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)`

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

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

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

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

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

\(=\left[\frac{2xy}{\left(x-y\right).\left(x+y\right)}+\frac{x-y}{2.\left(x+y\right)}\right]:\frac{x+y}{2x}+\frac{x}{y-x}\)

\(=\frac{4xy+\left(x-y\right).\left(x-y\right)}{2.\left(x-y\right).\left(x+y\right)}.\frac{2x}{x+y}+\frac{x}{y-x}\)

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

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

\(=\frac{x}{x-y}-\frac{x}{x-y}=0\)

Bạn giùm mik nhé, tks bạn nhiều (:

sai rồi

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 $x=-\dfrac32$.

giúp mk với nhé

sáng mai nộp rồi 

ai nhanh tay mk sẽ k cho

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

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

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

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

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

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