Question

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

Chú ý \(2x^3+x^2+2x+1=\left(2x+1\right)\left(x^2+1\right)\)

các bạn giúp mình nhé

Answer 1

\(VT\ge0;\forall x\Rightarrow VP\ge0\Rightarrow x\ge-\frac{1}{2}\)

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

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

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

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

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