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đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)
\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)
TH1: \(x-1=0\Leftrightarrow x=1\)
TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
ĐKXĐ:...
\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}=\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{-3x^2-2x+1}{3x}\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)\left(1-3x\right)}{3x\left(x+1\right)}\right].\frac{x}{x-1}=\left(\frac{2}{3x}-\frac{2\left(1-3x\right)}{3x}\right).\left(\frac{x}{x-1}\right)\)
\(=\left(\frac{2-2+6x}{3x}\right)\left(\frac{x}{x-1}\right)=\frac{2x}{x-1}\)
1) \(VT=x^3+x^2y-x^2y-xy^2+xy^2+y^3=x^3+y^3=VP\)
2) \(VP=x^2+xy-xy-y^2=x^2-y^2=VT\)
3) \(VP=x^2+2\cdot x\cdot1+1=x^2+2x+1=VT\)
4) \(VP=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3=VT\)
1, \(\left(x^2-xy+y^2\right)\left(x+y\right)=x^3+y^3\\ x^3+x^2y-x^2y-xy^2+xy^2+y^3=x^3+y^3\\ x^3+y^3=x^3+y^3\left(đúng\right)\)Vậy ta được đpcm
2, \(x^2-y^2=\left(x-y\right)\left(x+y\right)\\ x^2-y^2=x^2+xy-xy-y^2\\ x^2-y^2=x^2-y^2\left(đúng\right)\)Vậy ta được đpcm
3, \(x^2+2x+1=\left(x+1\right)^2\\ x^2+2x+1=\left(x+1\right)\left(x+1\right)\\ x^2+2x+1=x^2+x+x+1\\ x^2+2x+1=x^2+2x+1\left(đúng\right)\)Vậy ta được đpcm
4, \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\\ x^3-y^3=x^3+x^2y+xy^2-x^2y-xy^2-y^3\\ x^3-y^3=x^3-y^3\left(đúng\right)\)Vậy ta được đpcm
ta có : \(VP=x^3+3x^2+2x=x\left(x^2+3x+2\right)=x\left(x^2+x+2x+2\right)\)
\(=x\left(x\left(x+1\right)+2\left(x+1\right)\right)=x\left(x+2\right)\left(x+1\right)=VT\)
vậy \(x\left(x+1\right)\left(x+2\right)=x^3+3x^2+2x\) (đpcm)
Ta có \(VT\) :
\(x\left(x+1\right)\left(x+2\right)=x^3+2x^2+x^2+2x=x^3+3x^2+2x=VP\)
\(\Rightarrowđpcm\)