giải pt : $\frac{3}{\sqrt{3x-8}+1}$ – $\frac{1}{\sqrt{x+1}+2}$= $\frac{2}{5}$

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1. Đáp án:

   $x = 8$

   Giải thích các bước giải:

   $\begin{array}{l}\dfrac{3}{\sqrt{3x – 8} +1} – \dfrac{1}{\sqrt{x + 1} + 2} = \dfrac{2}{5}\qquad (*)\\ ĐKXĐ: \, x \geq \dfrac{8}{3}\\ (*) \Leftrightarrow \dfrac{3}{\sqrt{3x-8}+1}-\dfrac{3}{5}=\dfrac{1}{\sqrt{x+1}+2}-\dfrac{1}{5}\\ \Leftrightarrow \dfrac{3(4-\sqrt{3x-8})}{\sqrt{3x-8}+1}=\dfrac{3-\sqrt{x+1}}{\sqrt{x+1}+2}\\ \Leftrightarrow (8-x)\left[\dfrac{9}{(4+\sqrt{3x-8})(1+\sqrt{3x-8})}-\dfrac{1}{(3+\sqrt{x+1})(2+\sqrt{x+1})}\right]=0\\ \Leftrightarrow \left[\begin{array}{l}x =8\\\dfrac{9}{(4+\sqrt{3x-8})(1+\sqrt{3x-8})}-\dfrac{1}{(3+\sqrt{x+1})(2+\sqrt{x+1})} = 0 \quad (**)\end{array}\right.\\ (**) \Leftrightarrow 9(3+\sqrt{x+1})(2+\sqrt{x+1}) = (4+\sqrt{3x-8})(1+\sqrt{3x-8})\\ Ta \,\,có:\\ 9(3+\sqrt{x+1})(2+\sqrt{x+1})\\ = (3\sqrt3+\sqrt{3x+3})(6\sqrt3+3\sqrt{3x+3})> (4+\sqrt{3x-8})(1+\sqrt{3x-8})\\ \text{Do đó}\,\,(**)\,\,\text{vô nghiệm}\\ \text{Vậy phương trình có nghiệm duy nhất x = 8}\end{array}$

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