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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^2\left(1-x^2\right)-4-4x^2\)
\(=-x^4+x^2-4-4x^2\)
\(=-\left(x^4+4+4x^2-x^2\right)\)
\(=-\left(\left(x^2+2\right)^2-x^2\right)\)
\(=-\left(x^2+2-x\right)\left(x^2+2+x\right)\)
\(x^2\left(1-x^2\right)-4-4x^2=x^2\left(1-x\right)\left(1+x\right)-4\left(1+x^2\right)\)
Đến đấy tách thế nào đây ( đề sai hả )
\(x^3+4x^2+4x+1\)
\(=x^3+3x^2+x+x^2+3x+1\)
\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x+1\right)\)
a). 4x^2 - 9y^2 + 4x - 6y
=(2x-3y)(2x+3y)+2(2x-3y)
=(2x-3y)(2x+3y+2)
b) x(x+1)^2 + x(x-5) - 5(x+1)^2
=(x+1)2(x-5)+x(x-5)
=(x-5)[(x+1)2+x]
=(x-5)(x2+2x+1+x)
=(x-5)(x2+3x+1)
\(\left(x^2+x-1\right)^2+4x^2+4x-1\)
\(=\left(x^2+x-1\right)^2+4\left(x^2+x-1\right)+3\)
\(=\left(x^2+x-1\right)^2+x^2+x-1+3\left(x^2+x-1\right)+3\)
\(=\left(x^2+x-1\right)\left(x^2+x-1+1\right)+3\left(x^2+x-1+1\right)\)
\(=\left(x^2+x-1\right)\left(x^2+x\right)+3\left(x^2+x\right)\)
\(=\left(x^2+x\right)\left(x^2+x-1+3\right)\)
\(=x\left(x+1\right)\left(x^2+x+2\right)\)
Chúc bạn học tốt.
\(x^4+4x^3+2x^2-4x+1\)
\(=x^4+2x^3-x^2+2x^3+4x^2-2x-x^2-2x+1\)
\(=x^2\left(x^2+2x-1\right)+2x\left(x^2+2x-1\right)-\left(x^2+2x-1\right)\)
\(=\left(x^2+2x-1\right)^2\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(\left(x^2+1\right)^2-4x^2=\left(x^2+1\right)^2-\left(2x\right)^2=\left(x^2+1-2x\right)\left(x^2+1+2x\right)=\left(x-1\right)^2\left(x-1\right)^2\)
(x2+1)2 - 4x2
=(x2+1)2 - (2x)2
=(x2+1-2x)(x2+1+2x)
=(x2-2x+1)(x2+2x+1)
=(x-1)2(x+1)2