Question

giải phương trình vô tỉ sau

\(\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1\)

Answer 1

\(\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1\)

\(pt\Leftrightarrow\sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}\)

\(\Leftrightarrow\sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)\)

\(\Leftrightarrow\frac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}=\frac{2x^3-3x^2-3x+2}{4x-1}\)

\(\Leftrightarrow\frac{x^3-x^2-2x}{\sqrt{x^3+1}+x+1}-\frac{2x^3-3x^2-3x+2}{4x-1}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)\left(x+1\right)}{\sqrt{x^3+1}+x+1}-\frac{\left(x+1\right)\left(x-2\right)\left(2x-1\right)}{4x-1}=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}\right)=0\)

Suy ra x=2;x=-1 còn 1 nghiệm nữa xấu quá t gg :v