Question

giải phương trình

\(\sqrt{x^2-2x+5}+\sqrt{x^2+2x+10}=\sqrt{29}\)

Answer 1

Lời giải:

ĐK: \(x\in\mathbb{R}\)

\(\sqrt{x^2-2x+5}+\sqrt{x^2+2x+10}=\sqrt{29}\)

\(\Leftrightarrow \sqrt{x^2+2x+10}=\sqrt{29}-\sqrt{x^2-2x+5}\)

Bình phương 2 vế:

\(x^2+2x+10=29+x^2-2x+5-2\sqrt{29(x^2-2x+5)}\)

\(\Leftrightarrow 4x-24=-2\sqrt{29(x^2-2x+5)}\)

\(\Leftrightarrow 12-2x=\sqrt{29(x^2-2x+5)}\)

\(\Rightarrow \left\{\begin{matrix} 12-2x\geq 0\\ (12-2x)^2=29(x^2-2x+5)\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 6\\ 4x^2+144-48x=29x^2-58x+145\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 6\\ 25x^2-10x+1=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\leq 6\\ (5x-1)^2=0\end{matrix}\right.\Rightarrow x=\frac{1}{5}\)

(thỏa mãn)

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