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[(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

`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 $x=-\dfrac32$.

a) (x^2+2xy+y^2) : (x+y)

=(x+y)²:(x+y)

=x+y

b) (125x^3+1) : (5x+1)

=(5x+1)(25x²-5x+1):(5x+1)

=25x²-5x+1

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

=(x-y)²:(y-x)

=-(x-y)²:(x-y)

=-(x-y)

=-x+y

`@` `\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`

=(x-y)³:(x-y)²

=x-y

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

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

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

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

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

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