Question

giải phương trình: \(\sqrt{x^2+7}-\sqrt{x^2-5}=x-1\)

Answer 1

Lời giải:
ĐKXĐ: \(x^2\geq 5\)

PT \(\Leftrightarrow (\sqrt{x^2+7}-4)-(\sqrt{x^2-5}-2)=x-3\)

\(\Leftrightarrow \frac{x^2+7-16}{\sqrt{x^2+7}+4}-\frac{x^2-5-4}{\sqrt{x^2-5}+2}=x-3\)

\(\Leftrightarrow \frac{(x-3)(x+3)}{\sqrt{x^2+7}+4}-\frac{(x-3)(x+3)}{\sqrt{x^2-5}+2}=x-3\)

\(\Leftrightarrow (x-3)\left[1+\frac{x+3}{\sqrt{x^2-5}+2}-\frac{x+3}{\sqrt{x^2+7}+4}\right]=0(1)\)

Với \(\forall x^2\geq 5\) thì:

\(\left\{\begin{matrix} x+3>0\\ \sqrt{x^2-5}+2< \sqrt{x^2+7}+4\end{matrix}\right.\Rightarrow \frac{x+3}{\sqrt{x^2-5}+2}>\frac{x+3}{\sqrt{x^2+7}+4}\)

\(\Rightarrow 1+\frac{x+3}{\sqrt{x^2-5}+2}-\frac{x+3}{\sqrt{x^2+7}+4}\neq 0(2)\)

Từ (1);(2) \(\Rightarrow x-3=0\Rightarrow x=3\) (thỏa mãn)

Vậy.......