=\(\left(3x^2-4x-13-4x^2+9\right)\left(3x^2-4x-13+4x^2-9\right)-\left(x+2\right)^4\)

=\(\left(-x^2-4x-4\right)\left(7x^2-4x-22\right)-\)\(\left(x+2\right)^{^{ }2.2}\)

=\(-\left(x+2\right)^2\left(7x^2-4x-22\right)-\left(x+2\right)^2\left(x+2\right)^2\)

=\(-\left(x+2\right)^2\)\(\left(7x^2-4x-22-x^2-4x-4\right)\)

\(-\left(x+2\right)^2\)(\(6x^2-8x-26\))

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

a)x(x^2-x+2)(x^2+x+2)

còn lại thì vào đây: https://coccoc.com/search/math

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

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

`9(x-y)^2-4(x+y)^2`

`=[3(x-y)]^2-[2(x+y)]^2`

`=(3x-3y)^2-(2x+2y)^2`

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

`=(5x-y)(x-5y)`

\(9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)

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

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

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

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

\(x^4+8x^3+28x^2+48x-13\)

\(=x^4+4x^3+13x^2+4x^3+16x^2+52x-x^2-4x-13\)

\(=x^2\left(x^2+4x+13\right)+4x\left(x^2+4x+13\right)-\left(x^2+4x+13\right)\)

\(=\left(x^2+4x-1\right)\left(x^2+4x+13\right)\)

Ta có:\(9\left(x+5\right)^2-4\left(x-7\right)^2=9\left(x^2+10x+25\right)-4\left(x^2-14x+49\right)\)

\(=9x^2+90x+225-4x^2+56x-196=5x^2+146x+29\)

\(=\left(5x^2+145x\right)+\left(x+29\right)=\left(x+29\right)\left(5x+1\right)\)

Bài làm:

Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)

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

Học tốt!!!!

Ta có : 

\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)

\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)