Question

tìm x:x^3-6x^2+12x+19=0

Answer 1

\(=\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=\left(x+1\right)\left(x^2-7x+19\right)=0\)

Ta thấy:  \(x^2-7x+19=x^2-2\times\frac{7}{2}x+\frac{7}{2}^2+\frac{27}{4}=\left(x-\frac{7}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)lớn hơn 0

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

Answer 2

\(x^3-6x^2+12x+19=0\)

\(\Leftrightarrow\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(19x+19\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-7x+19\right)=0\)

Mà \(x^2-7x+19>0\)với \(\forall x\)

\(\Rightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy \(x=-1\)