4x²-49-12xy+9y²

=[(2x)²-2.2x.3y+(3y)²] - 7²

=(2x-3y)² - 7²

=(2x-3y - 7) ( 2x-3y+7)

a) \(4x^2-49-12xy+9y^2\)

\(< =>-\left(4x^2+12xy+9y^2\right)-49\)

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

a: \(ab+a+b+1\)

\(=a\left(b+1\right)+\left(b+1\right)\)

\(=\left(b+1\right)\left(a+1\right)\)

c: \(4x^2-12xy+3x-9y\)

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

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

a)x²-4y²

=x²-(2y)²

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

b)x³+27y³

=x³+(3y)³

=(x+3y)(x²-3xy+9y²)

c)4x²+12xy+9y²-16

=(2x+3y)²-4²

=(2x+3y-4)(2x+3y+4)

d)9x²-24xy+16y²-64

=(3x-4y)²-8²

=(3x-4y-8)(3x-4y+8)

e)8x³-27y³

=(2x)³-(3y)³

=(2x-3y)(4x²+6xy+9y²)

f)5x³-7x²+10x-14

=5x³+10x-7x²-14

=5x(x²+2)-7(x²+2)

=(5x-7)(x²+2)

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a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

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

c) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-25\right]\)

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

d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)

\(=\left(x-y\right)^2-\left(2m-n\right)^2\)

\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)

a) (4x²-3x-18)²-(4x²+3x)²

=(4x²-3x-18-4x²-3x)(4x²-3x-18+4x²+3x)

=(-6x-18)(8x²-18)

=-48x³+108x-144x²+324

a/25-9y^2-4x^2+12xy

=-(9y^2-12xy+4x^2-25)

=-[(3y)^2-2.3y.2x+(2x)^2-5^2]

=-[(3y-2x)^2-5^2]

=-(3y-2x-5)(3y-2x+5)

b/4x^2-8xy+4y^2-5x+5y

=4(x^2-2.x.y+y^2)-5(x-y)

=4(x-y)^2-5(x-y)

=(x-y)(4x-4y-5)

a: =(2x-3y)^2-4(2x-3y)

=(2x-3y)(2x-3y-4)

b: =3x^2+21x-x-7

=(x+7)(3x-1)

c: =(3x-1)^4+2(3x-1)^2+1

=[(3x-1)^2+1]^2

d: =2x^3-2x^2-x^2+x+x-1

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

\(a,4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

\(b,x^2-9x+20=x^2-4x-5x+20\)

\(=x\left(x-4\right)-5\left(x-4\right)\)

\(=\left(x-4\right)\left(x-5\right)\)

\(c,x^2+7x+12=x^2+3x+4x+12\)

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

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

\(1,=8xy+14y^2-4xz-7yz\\ 2,=y\left(4x^2-12x+9\right)=y\left(2x-3\right)^2\\ 3,\Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Câu 1: \(\left(2y-z\right)\left(4x+7y\right)=8xy-4xz+14y^2-7yz\)

câu 2: \(4x^2y-12xy+9y=y\left(4x^2-12x+9\right)\)

câu 3: \(\left(x-2\right)\left(x+3\right)+x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\\ \Leftrightarrow2\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)