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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3.\left(x-2\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
\(=x\left(x-6\right)+2\left(x-6\right)=\left(x-6\right)\left(x+2\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)
\(=\left[\left(x^2+1\right)-3\right]^2\)
\(=\left(x^2+1-3\right)^2\)
\(=\left(x^2-2\right)^2\)
\(x^2+x-2\)
\(=x^2-x+2x-2\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
chì làm được 1 cách thôi
Diệu Linh_face
\(6x^2+x-2\)
\(=6x^2-3x+4x-2\)
\(=3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(3x+2\right)\)
thak nha