`a, -xy(x^2+xy-y^2)`

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

`b, 5x^2y(2y^2-xy)`

`= 10x^2y^3 - 5x^3y^2`.

`c, (-2x^3 - 1/4y - 4y^2).8xy^2`.

`= -16x^4y^2 - 2xy^3 - 32xy^4`.

`d, (2x^3 - 3xy + 12x)(-1/6xy)`

`= -2/3x^4y + 1/2x^2y^2 - 2x^2y`.

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: \(\frac{4x^5y^2+2x^4y-6x^3y^2+3xy^4-y^5}{2x^2+xy-y^2}\)

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

Tính ra nữa được k ạ @@

\(3x^2\left(2x^3-3xy+4\right)\)

\(=3x^2.2x^3-3x^2.3xy+3x^2.4\)

\(=6x^5-9x^3y+12x^2\)

Di in anh ma dc tick la sao ?

`#3107.101107`

a)

`A + B =` \(x^2+5xy-3y^2\)\(+ 2x^2-3xy+11y^2\)

`= (x^2 + 2x^2) + (5xy - 3xy) + (-3y^2 + 11y^2)`

`= 3x^2 + 2xy + 8y^2`

b)

\((9x^3y^2-12x^2y+15xy) \div (3xy)\)

`= 9x^3y^2 \div 3xy - 12x^2y \div 3xy + 15xy \div 3xy`

`= 3x^2y - 4x + 5`