\(2x^3-3x^2+3x-1=0\)
\(\Leftrightarrow2x^3-2x^2-x^2+2x+x-1=0\)
\(\Leftrightarrow\left(2x^3-2x^2+2x\right)-\left(x^2-x+1\right)=0\)
\(\Leftrightarrow2x\left(x^2-x+1\right)-\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-x+1\right)=0\)
\(TH1:2x-1=0\Leftrightarrow x=\frac{1}{2}\)
\(TH2:x^2-x+1=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\)
Mà \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)nên loại TH2
Vậy \(x=\frac{1}{2}\)
2x3 - 3x2 + 3x - 1 = 0
(2x - 1)(x2 - x + 1) = 0
Vì: x2 - x + 1 > 0 nên:
2x - 1 = 0
2x = 0 + 1
2x = 1
x = 1/2
\(2x^3-3x^2+3x-1=0\)
\(\Rightarrow\left(x-1\right)^3+x^3=0\)
\(\Rightarrow\left(2x-1\right)\left(x^2-2x+1-x^2+x+x^2\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(x^2-x+1\right)=0\)
Vì \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\forall x\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\frac{1}{2}\)
