\(x^2-xz-9y^2+3yz\)

\(=\)\(\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

Chúc bạn học tốt ~ 

\(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

Thắng ơi bạn sai đề rồi nhé
Đề đúng: x2-xy-9y2+3yz
= (x² - 9y²) - (xz - 3yz) = (x - 3y)(x + 3y) - z(x - 3y) 
= (x - 3y)( x + 3y - z) 
 

đề mi khác gì ?

a) \(2x^3+6xy-x^2z-3yz\)

= \(\left(2x^3+6xy\right)-\left(x^2z+3yz\right)\)

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

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

b)\(x^2-6xy+9y^2-49\)

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

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

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

Cứu với mn ơi , mai em có ktra ạ 

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

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

c) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)

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

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

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

b) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(5+x\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

5xy³ + 30x²z² - 6x3yz - 25x²z

  => x(5y³ + 30xz² - 6x²yz - 25xz)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}