Bài 3:
a: \(20x-5y=5\left(4x-y\right)\)
b: \(5x\left(x-1\right)-3x\left(x-1\right)=2x\left(x-1\right)\)
c: \(x\left(x+y\right)-6x-6y=\left(x+y\right)\left(x-6\right)\)
d: \(6x^3-9x^2=3x^2\left(2x-3\right)\)
e: \(4x^2y-8xy^2+10x^2y^2=2xy\left(2x-5y+4xy\right)\)
f: \(20x^2y-12x^3=4x^2\left(5y-3x\right)\)
g: \(8x^4+12x^2y^4-16x^3y^4=4x^2\left(2x^2+3y^4-4xy^4\right)\)
h: \(4xy^2+8xyz=4xy\left(y+2z\right)\)
Bài 1:
a. $x^3+3x=x(x^2+3)$
b. $4x-8y=4(x-2y)$
c. $8(x+3y)-16x(x+3y)=(x+3y)(8-16x)=8(x+3y)(1-2x)$
d. $3(x-y)-5x(y-x)=3(x-y)+5x(x-y)=(x-y)(3+5x)$
e. $9x^2-6x=3x(3x-2)$
f.
$x^4y-2x^2y^2+5xy=xy(x^3-2xy+5)$
g. $4x^2(x+1)+2x^2(x+1)=(x+1)(4x^2+2x^2)$
$=6x^2(x+1)$
h.
$\frac{4}{3}x(y-2)-\frac{2}{5}y(2-y)=\frac{4}{3}x(y-2)+\frac{2}{5}y(y-2)$
$=(y-2)(\frac{4}{3}x+\frac{2}{5}y)$
$=2(y-2)(\frac{2}{3}x+\frac{1}{5}y)$
Bài 2:
a.
$2(x-1)^3-5(x-1)^2-(x-1)=(x-1)[2(x-1)^2-5(x-1)-1]$
$=(x-1)(2x^2-9x+6)$
b.
$y(y-x)^3-x(x-y)^2+xy(x-y)=-y(x-y)^3-x(x-y)^2+xy(x-y)$
$=(x-y)[-y(x-y)^2-x(x-y)+xy]$
$=-(x-y)(x^2y+x^2-2xy^2-2xy+y^3)$
Bài 3:
a. $20x-5y=5(4x-y)$
b. $5x(x-1)-3x(x-1)=(x-1)(5x-3x)=2x(x-1)$
c. $x(x+y)-6x-6y=x(x+y)-6(x+y)=(x+y)(x-6)$
d.
$6x^3-9x^2=3x^2(2x-3)$
e. $4x^2y-8xy^2+10x^2y^2$
$=2xy(2x-4y+5xy)$
f. $20x^2y-12x^3=4x^2(5y-3x)$
g. $8x^4+12x^2y^4-16x^3y^4=4x^2(2x^2+3y^4-4xy^4)$
h.
$4xy^2+8xyz=4xy(y+2z)$
Bài 4:
a. $3x(x+1)-5y(x+1)=(x+1)(3x-5y)$
b. $3x(x-6)-2(x-6)=(x-6)(3x-2)$
c. $4y(x-1)-(1-x)=4y(x-1)+(x-1)=(x-1)(4y+1)$
d. $(x-3)^3+3-x=(x-3)^3-(x-3)=(x-3)[(x-3)^2-1]$
$=(x-3)(x-3-1)(x-3+1)=(x-3)(x-4)(x-2)$
e.
$7x(x-y)-(y-x)=7x(x-y)+(x-y)=(x-y)(7x+1)$
Bài 4:
f. $3x^3(2y-3z)-15x(2y-3z)^2$
$=3x(2y-3z)[x^2-5(2y-3z)]=3x(2y-3z)(x^2-10y+15z)$
g.
$3x(z+2)+5(-z-2)=3x(z+2)-5(z+2)=(z+2)(3x-5)$
h.
$18x^2(3+x)+3(x+3)=3(x+3)(6x^2+1)$
i.
$14x^2y-21xy^2+28x^2y^2=7xy(2x-3y+4xy)$
k.
$10x(x-y)-8y(y-x)=10x(x-y)+8y(x-y)=(x-y)(10x+8y)$
$=2(x-y)(5x+4y)$
Bài 5:
a. $4x(x+1)=8(x+1)$
$\Leftrightarrow 4x(x+1)-8(x+1)=0$
$\Leftrightarrow (x+1)(4x-8)=0$
$\Leftrightarrow x+1=0$ hoặc $4x-8=0$
$\Leftrightarrow x=-1$ hoặc $x=2$
b.
$x(x-1)-2(1-x)=0$
$\Leftrightarrow x(x-1)+2(x-1)=0$
$\Leftrightarrow (x+2)(x-1)=0$
$\Leftrightarrow x+2=0$ hoặc $x-1=0$
$\Leftrightarrow x=-2$ hoặc $x=1$
c.
$2x(x-2)-(2-x)^2=0$
$\Leftrightarrow 2x(x-2)-(x-2)^2=0$
$\Leftrightarrow (x-2)(2x-x+2)=0$
$\Leftrightarrow (x-2)(x+2)=0$
$\Leftrightarrow x-2=0$ hoặc $x+2=0$
$\Leftrightarrow x=\pm 2$
d.
$(x-3)^3+3-x=0$
$\Leftrightarrow (x-3)^2-(x-3)=0$
$\Leftrightarrow (x-3)(x-3-1)=0$
$\Leftrightarrow (x-3)(x-4)=0$
$\Leftrightarrow x-3=0$ hoặc $x-4=0$
$\Leftrightarrow x=3$ hoặc $x=4$
e.
$5x(x-2)-(2-x)=0$
$\Leftrightarrow (x-2)(5x+1)=0$
$\Leftrightarrow x-2=0$ hoặc $5x+1=0$
$\Leftrightarrow x=2$ hoặc $x=-\frac{1}{5}$
Em ơi em tách nhỏ bài ra để hỏi nha ^^
Bài 2:
a: Ta có: \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)
\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)
\(=\left(x-1\right)\left(2x^2-9x+6\right)\)
b: Ta có: \(-y\left(x-y\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)\)
\(=\left(x-y\right)\left[-y\left(x-y\right)^2-x\left(x-y\right)+xy\right]\)
\(=\left(x-y\right)\left(-x^2y+2xy^2-y^3-x^2+xy+xy\right)\)
\(=\left(x-y\right)\left(-x^2y+2xy^2-y^3-x^2+2xy\right)\)
