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12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 8 2021
\(=\left(x^2+y^2-x^2y^2-1\right)+\left(xy-x-y+1\right)\)
\(=\left(x^2-1\right)-y^2\left(x^2-1\right)+x\left(y-1\right)-\left(y-1\right)\)
\(=\left(x^2-1\right)\left(1-y^2\right)+\left(x-1\right)\left(y-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(1-y\right)\left(1+y\right)-\left(x-1\right)\left(1-y\right)\)
\(=\left(x-1\right)\left(1-y\right)\left[\left(x+1\right)\left(y+1\right)-1\right]\)
\(=\left(x-1\right)\left(1-y\right)\left(xy+x+y\right)\)
9 tháng 11 2018
a) \(xy-y^2-x+y\)
\(=y\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(y-1\right)\)
b) \(xy-x^2-x+y\)
\(=x\left(y-x\right)+\left(y-x\right)\)
\(=\left(y-x\right)\left(x+1\right)\)
BD
1
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
N
0
N
1
8 tháng 10 2021
b: \(3x+3y-x^2-2xy-y^2\)
\(=3\left(x+y\right)-\left(x+y\right)^2\)
\(=\left(x+y\right)\left(3-x-y\right)\)
\(y-x^2y-xy^2-y^2\)
\(=\left(y-x^2y\right)-\left(xy^2+y^2\right)\)
\(=y\left(1-x^2\right)-y^2\left(x+1\right)\)
\(=y\left(1-x\right)\left(x+1\right)-y^2\left(x+1\right)\)
\(=\left(x+1\right)\left[y\left(1-x\right)-y^2\right]\)
\(=\left(x+1\right)\left(y-xy-y^2\right)\)
\(=\left(x+1\right)\left[y-\left(xy+y^2\right)\right]\)
\(=y\left(x+1\right)\left(1-x-y\right)\)