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

a)x³+3x²+3x+1-27z³

=(x+1)³-(3z)³

=(x+1-3z)[(x+1)²+3z(x+1)+9z²

b)81x⁴+4

=(9x²)²+2²

=(9x²+2)²-(6x)²

=(9x²-6x+2)(9x²+6x+2)

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

= x(x - 1 - 2y)(x - 1 + 2y)

Bài 3:

a: \(2x\left(x-3\right)-x+3=0\)

=>\(2x\left(x-3\right)-\left(x-3\right)=0\)

=>(x-3)(2x-1)=0

=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

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

=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)

=>\(5x^2-x-2=5x^2\)

=>-x-2=0

=>-x=2

=>x=-2

pls help me mk đang cần vội :(

Bài 1:

\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)

Bài 2:

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

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

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)

\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

ỳtct7ct7c7c7t79tc9

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

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

           Bài làm :

 \(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)

\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

 \(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)

\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)

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

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

 \(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

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

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

\(d ) x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3