\(=-2x^2+6xy-3y^2+25\)

\(=-\left(2x^2-6xy+3y^2\right)+25\)

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

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

\(=\left(2x^2+3y^2\right).\left(2x^2-3y^2\right)+25\)

quên rùi

\(-4x^2+12xy-9y^2+25=\left(-4x^2+6xy+10x\right)+\left(6xy-9y^2-15y\right)-\left(10x-15y-25\right)\)

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

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

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: \(ab+a+b+1\)

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

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

c: \(4x^2-12xy+3x-9y\)

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

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

1. (x-3)²

2. (3y+2x)²

3. (1/5x-8y)(1/5x+8y)

4. (x-2y)(x²+2xy+4y²)

5. (4x-3-x-1)(4x-3+x+1)

(3x-4)(5x-2)

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: =(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)

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

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