Question

bài 2:giải các phương trình sau
a,x^3+x^2+x=-1/3
b,x^3+x^2-x+2=0

Answer 1

a.

\(\Leftrightarrow3x^3+3x^2+3x=-1\)

\(\Leftrightarrow x^3+3x^2+3x+1=-2x^3\)

\(\Leftrightarrow\left(x+1\right)^3=\left(-\sqrt[3]{2}x\right)^3\)

\(\Leftrightarrow x+1=-\sqrt[3]{2}x\)

\(\Leftrightarrow\left(1+\sqrt[3]{2}\right)x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{1+\sqrt[3]{2}}\)

b.

\(\Leftrightarrow x^3-x^2+x+2x^2-2x+2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\Rightarrow x=-2\\x^2-x+1=0\left(vn\right)\end{matrix}\right.\)

Answer 2

b) Ta có: \(x^3+x^2-x+2=0\)

\(\Leftrightarrow x^3+2x^2-x^2-2x+x+2=0\)

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

\(\Leftrightarrow x+2=0\)

hay x=-1