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\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`
`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`
`c, xy^2 + x^2y + 1/4y^3`
`= y(xy + x^2 + 1/4y^2)`
`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`
`= (x+1+y)(x+1-y)`
1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(a,x^2+9x+20=x^2+4x+5x+20.\)
\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(c,x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2-2\right)-\left(2x\right)^2=\left(x^2-2x-2\right)\left(x^2+2x-2\right)\)
\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
1) \(3x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\)
2) \(4x(x-2y)-8y(2y-x)\)
\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)
\(=\left(4x+8y\right)\left(x-2y\right)\)
\(=4\left(x+2y\right)\left(x-2y\right)\)
3) \(a^2\left(x-1\right)+b^2\left(1-x\right)\)
\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)
\(=\left(a^2-b^2\right)\left(x-1\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(x-1\right)\)
4) \(3x\left(x-a\right)+4a\left(a-x\right)\)
\(=3x\left(x-a\right)-4a\left(x-a\right)\)
\(=\left(x-a\right)\left(3x-4a\right)\)
5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)
\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)
\(=\left(5x+10y^2\right)\left(x-y\right)^2\)
\(=5\left(x+2y^2\right)\left(x-y\right)^2\)
6) \(3x\left(x-3\right)^2+9\left(3-x\right)^2\)
\(=3x\left(x-3\right)^2+9\left(x-3\right)^2\)
\(=\left(3x+9\right)\left(x-3\right)^2\)
\(=3\left(x+3\right)\left(x-3\right)^2\)
7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)
\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)
\(=\left(x-y\right)\left(a-m\right)^2\)
8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)
\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)
\(=\left(6y^2+9x\right)\left(x-1\right)^2\)
\(=3\left(2y^2+3x\right)\left(x-1\right)^2\)
#Ayumu
2x3+3x2+8x-5=2x3-x2+4x2-2x+10x-5
=x2.(2x-1)+2x.(2x-1)+5.(2x-1)
= (2x-1).(x2+2x+5)
Ta có: