Tìm x biết
\(x^3+\left(x-1\right)^3=\left(2x-1\right)^3\)
\(\Leftrightarrow x^3+\left(x-1\right)^3-\left(2x-1\right)^3=0\)
\(\Leftrightarrow x^3+x^3-3x^2+3x-1-8x^3+12x^2-6x+1=0\)
\(\Leftrightarrow-6x^3+9x^2-3x=0\)
\(\Leftrightarrow6x^3-9x^2+3x=0\)
\(\Leftrightarrow3x\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow3x\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow3x\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;1;\frac{1}{2}\right\}\)
Ta có : \(x^3+\left(x-1\right)^3=\left(2x-1\right)^3\)
=> \(x^3=\left(2x-1\right)^3-\left(x-1\right)^3\)
=> \(x^3=\left(2x-1-x+1\right)\left(\left(2x-1\right)^2+\left(2x-1\right)\left(x-1\right)+\left(x-1\right)^2\right)\)
=> \(x^3=x\left(4x^2-4x+1+2x^2-x-2x+1+x^2-2x+1\right)\)
=> \(x^3-x\left(7x^2-9x+3\right)=0\)
=> \(x\left(x^2-7x^2+9x-3\right)=0\)
=> \(x\left(2x^2-3x+1\right)=0\)
=> \(x\left(2x^2-2x-x+1\right)=0\)
=> \(x\left(2x\left(x-1\right)-\left(x-1\right)\right)=0\)
=> \(x\left(x-1\right)\left(2x-1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x-1=0\\2x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy ...
