\(a\)) \(x^3-3x^2+4x-2=0\)
\(\Leftrightarrow x^3-x^2-2x^2+2x+2x-2=0\)
\(\Leftrightarrow x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy phương trình có 1 nghiệm \(x\in\left\{1\right\}\)
\(b\)) \(x^2-2x-5=0\)
\(\Leftrightarrow x^2-\left(1-\sqrt{6}\right)x-\left(1+\sqrt{6}\right)x-5=0\)
\(\Leftrightarrow\left[x-\left(1+\sqrt{6}\right)\right].\left[x-\left(1-\sqrt{6}\right)\right]=0\)
Suy ra \(x-\left(1+\sqrt{6}\right)=0\) hoặc \(x-\left(1-\sqrt{6}\right)=0\)
\(x-\left(1+\sqrt{6}\right)=0\) \(\Leftrightarrow x=1+\sqrt{6}\)\(x-\left(1-\sqrt{6}\right)=0\) \(\Leftrightarrow x=1-\sqrt{6}\)Vậy phương trình có 2 nghiệm \(x\in\left\{1-\sqrt{6};1+\sqrt{6}\right\}\)
