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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
a: \(=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b: \(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
c: \(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)\)
\(=\left(x-8\right)\left(x+1\right)\)
\(a,xy+y^2-x-y\)
\(=\left(xy+y^2\right)-\left(x+y\right)\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
\(---\)
\(b,\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
\(---\)
\(c,x^2-7x-8\)
\(=x^2+x-8x-8\)
\(=\left(x^2+x\right)-\left(8x+8\right)\)
\(=x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x-8\right)\)
\(Toru\)
\(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left(\frac{y-x}{xy}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{\left(x-y\right)^2}{x^2y^2}-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2x^2y^2}{xy\left(x-y\right)^2}-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2xy}{\left(x-y\right)^2}-\frac{x^2+y^2}{\left(x-y\right)^2}=\frac{-x^2+2xy-y^2}{\left(x-y\right)^2}\)
\(=-\frac{\left(x-y\right)^2}{\left(x-y\right)^2}=-1\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(A=\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2-\left(xy+yz+zx\right)^2\left(1\right)\)
Đặt \(x^2+y^2+z^2=a\)
\(xy+yz+zx=b\Rightarrow2\left(xy+yz+zx\right)=2b\)
\(\Rightarrow a+2b=\left(x+y+z\right)^2\)
Kết hợp (1) ta được : \(A=a\left(a+2b\right)+b^2\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)
a) \(xy+y^2-x-y\)
\(=\left(xy+y^2\right)-\left(x+y\right)\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
b) \(\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8\right)^2-1^2\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)
\(=\left(xy+3\right)\left(xy-3\right)\left(x^2y^2-7\right)\)
`a, xy +y^2 -x-y`
`=(xy+y^2)-(x+y)`
`= y(x+y)-(x+y)`
`= (x+y)(y-1)`
`b, (x^2y^2 -8)^2 -1`
`= (x^2y^2 -8)^2 -1^2`
`=(x^2y^2 -8-1)(x^2y^2-8+1)`
`=(x^2y^2 -9)(x^2y^2-7)`