Câu hỏi của asuna

Question

giải pt $x^2+x+\frac{1}{2}=\sqrt{2x^3-x^2+x+1}$

Khôi Bùi · Accepted Answer

$x^2+x+\frac{1}{2}=\sqrt{2x^3-x^2+x+1}$

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

$\Leftrightarrow x^4+x^2+\frac{1}{4}+2x^3+x^2+x=2x^3-x^2+x+1$

$\Leftrightarrow x^4+2x^3+2x^2+x+\frac{1}{4}=2x^3-x^2+x+1$

$\Leftrightarrow x^4+3x^2-\frac{3}{4}=0$

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

$\Leftrightarrow\left(x^2+\frac{3}{2}\right)^2=3$

$\Leftrightarrow\left[{}\begin{matrix}x^2+\frac{3}{2}=\sqrt{3}\x^2+\frac{3}{2}=-\sqrt{3}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1,5\x^2=-\sqrt{3}-\frac{3}{2}\left(VL\right)\end{matrix}\right.$

$\Leftrightarrow x=\pm\sqrt{\sqrt{3}-1,5}$

Vậy ...Câu trả lời: $x^2+x+\frac{1}{2}=\sqrt{2x^3-x^2+x+1}$