Tìm một nghiệm của đa thức f(x)=x^3-x^2+x-1

a)\(f\left(x\right)=x^3-x^2+x-1\)

Cho \(f\left(x\right)=0\Rightarrow x^3-x^2+x-1=0\)

\(\Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+1\right)=0\)

Dễ thấy: \(x^2+1\ge1>0\forall x\) ( vô nghiệm )

\(\Rightarrow x-1=0\Rightarrow x=1\)

b)\(g\left(x\right)=11x^3+5x^2+4x+10\)

Cho \(g\left(x\right)=0\Rightarrow11x^3+5x^2+4x+10=0\)

\(\Rightarrow11x^3-6x^2+10x+11x^2-6x+10=0\)

\(\Rightarrow x\left(11x^2-6x+10\right)+\left(11x^2-6x+10\right)=0\)

\(\Rightarrow\left(x+1\right)\left(11x^2-6x+10\right)=0\)

Dễ thấy:

\(11x^2-6x+10=11\left(x-\dfrac{3}{11}\right)^2+\dfrac{101}{11}\ge\dfrac{101}{11}>0\forall x\) (vô nghiệm)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

c)\(h\left(x\right)=-17x^3+8x^2-3x+12\)

Cho \(h\left(x\right)=0\Rightarrow-17x^3+8x^2-3x+12=0\)

\(\Rightarrow17x^2+9x+12-17x^3-9x^2-12x=0\)

\(\Rightarrow\left(17x^2+9x+12\right)-x\left(17x^2+9x+12\right)=0\)

\(\Rightarrow\left(1-x\right)\left(17x^2+9x+12\right)=0\)

Dễ thấy:

\(17x^2+9x+12=17\left(x+\dfrac{9}{34}\right)^2+\dfrac{735}{68}\ge\dfrac{735}{68}>0\forall x\)(vô nghiệm)

\(\Rightarrow1-x=0\Rightarrow x=1\)