Giúp với mọi người

c: \(=3x^3-4x^2+6x^2-8x=3x^2+2x^2-8x\)

a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)

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

\(=5x^3-7x^2y+2xy^2+5x-2y\)

b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

\(=\left(x^2-1\right)\left(x+2\right)\)

\(=x^3+2x^2-x-2\)

c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)

\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)

\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)

`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`.

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

\(=5x^3-7x^2y+2xy^2+5x-2y\)

thanks bạn nha

a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

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: \(\left(2x^4-x^3+5x^{}-6x^2-1\right):\left(1-2x\right)\)

\(=\frac{2x^4-x^3-6x^2+5x-1}{-2x+1}\)

\(=\frac{-2x^4+x^3+6x^2-5x+1}{2x-1}=\frac{-x^3\left(2x-1\right)+6x^2-3x-2x+1}{2x-1}\)

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

b: \(\left(x^3-3x^2+3x-2\right):\left(x^2-x+1\right)\)

\(=\frac{x^3-x^2+x-2x^2+2x-2}{x^2-x+1}\)

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

c: \(\left(2x^3+5x^2-2x+3\right):\left(2x^2-x+1\right)\)

\(=\frac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}\)

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

d: \(\frac{5xy^2}{x^2y}+\frac{5y-x^2}{xy^2}\)

\(=\frac{y\cdot5xy^2+x\left(5y-x^2\right)}{x^2y^2}=\frac{5x^2y^2+5xy-x^3}{x^2y^2}=\frac{x\left(5xy^2+5y-x^2\right)}{x^2y^2}\)

\(=\frac{5xy^2+5y-x^2}{xy^2}\)

e: \(\frac{27-x^3}{5x^5}:\frac{2x-6}{3x^3}\)

\(=\frac{-\left(x-3\right)\left(x^2+3x+9\right)}{5x^5}\cdot\frac{3x^3}{2\left(x-3\right)}=\frac{-3\left(x^2+3x+9\right)}{5x^2\cdot2}=\frac{-3\left(x^2+3x+9\right)}{10x^2}\)

Bài 3: 

a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)

b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)

c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)