a. 12xy² - 8x²y = 4xy . (3y - 2x)

b. 3x + 3y - x² - xy = (3x + 3y) - (x² + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)

GIUSP MIK VS MN ƠI

Bài 2:

a: =>2x^2-8+3x-2x^2=15

=>3x=23

=>x=23/3

b: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

=>x=5 hoặc x=-3

c: =>x^2+7x-30=0

=>(x+10)(x-3)=0

=>x=3 hoặc x=-10

a: \(=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)

b: \(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)

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

c: \(=x^2-8x+x-8\)

\(=x\left(x-8\right)+\left(x-8\right)\)

\(=\left(x-8\right)\left(x+1\right)\)

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

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

\(=y\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(y-1\right)\)

\(---\)

\(b,\left(x^2y^2-8\right)^2-1\)

\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)

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

\(---\)

\(c,x^2-7x-8\)

\(=x^2+x-8x-8\)

\(=\left(x^2+x\right)-\left(8x+8\right)\)

\(=x\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(x-8\right)\)

\(Toru\)

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

a: \(P=-3x^3+5x\)

\(=x\cdot\left(-3x^2\right)+x\cdot5\)

\(=x\left(-3x^2+5\right)\)

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

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

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

c: \(R=4-16x^2\)

\(=4\cdot1-4\cdot4x^2\)

\(=4\left(1-4x^2\right)\)

\(=4\left(1-2x\right)\left(1+2x\right)\)

d: \(S=36-4x^2\)

\(=4\cdot9-4\cdot x^2\)

\(=4\left(9-x^2\right)\)

\(=4\left(3-x\right)\left(3+x\right)\)

e: \(T=8x^3-1\)

\(=\left(2x\right)^3-1^3\)

\(=\left(2x-1\right)\left(4x^2+2x+1\right)\)

f: \(Q=8-x^3\)

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

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

g: \(N=64-x^3\)

\(=4^3-x^3\)

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

Bài 1:

a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$

$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$

b.

$(x+1)(x+2)(x+3)(x+4)-24$

$=[(x+1)(x+4)][(x+2)(x+3)]-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

$=a(a+2)-24$ (đặt $x^2+5x+4=a$)

$=a^2+2a-24=(a^2-4a)+(6a-24)$

$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$

$=x(x+5)(x^2+5x+10)$

Bài 2:

a. ĐKXĐ: $x\neq 3; 4$

\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)

b. $x^2+20=9x$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Rightarrow x=5$ (do $x\neq 4$)

Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$

Bài 3:

$(2x^2-7x^2:13x:2):(2x-1)=(2x^2-\frac{7}{26}x):(2x-1)$

$=[x(2x-1)+\frac{19}{52}(2x-1)+\frac{19}{52}]:(2x-1)$

$=[(2x-1)(x+\frac{19}{52})+\frac{19}{52}]: (2x-1)$

$\Rightarrow$ thương là $x+\frac{19}{52}$ và thương là $\frac{19}{52}$

Bài `1:`

`a)3x^3+6x^2=3x^2(x+2)`

`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`

Bài `2:`

`a)(2x-1)^2-25=0`

`<=>(2x-1-5)(2x-1+5)=0`

`<=>(2x-6)(2x+4)=0`

`<=>[(x=3),(x=-2):}`

`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`

`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`

`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`

`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`

Lời giải :

\(B=2bc\left(b+2c\right)+2ac\left(c-2a\right)-2ab\left(a+2b\right)-7abc\)

\(B=2b^2c+4bc^2+2ac^2-4a^2c-2ab\left(a+2b\right)-7abc\)

\(B=abc+2b^2c-4a^2c-8abc-2ab\left(a+2b\right)+2ac^2+4bc^2\)

\(B=bc\left(a+2b\right)-4ac\left(a+2b\right)-2ab\left(a+2b\right)+2c^2\left(a+2b\right)\)

\(B=\left(a+2b\right)\left(bc-4ac-2ab+2c^2\right)\)

\(B=\left(a+2b\right)\left[c\left(2c+b\right)-2a\left(2c+b\right)\right]\)

\(B=\left(a+2b\right)\left(2c+b\right)\left(c-2a\right)\)