Lời giải
{\begin{eqnarray*} && \sqrt{\dfrac{x-1}{x+1}}+\dfrac{2x+6}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}=2\\ &\Leftrightarrow& \sqrt{\dfrac{x-1}{x+1}}+\dfrac{2x+6}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}-2=0\\ &\Leftrightarrow& \sqrt{\dfrac{x-1}{x+1}}+2\left[\dfrac{x+3}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}-1\right]=0\\ &\Leftrightarrow& \sqrt{\dfrac{x-1}{x+1}} - 2\left[\dfrac{x-1+2\sqrt{(x-1)(x+3)}}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}\right]=0\\ &\Leftrightarrow& \sqrt{\dfrac{x-1}{x+1}} -2\left[\dfrac{\sqrt{x-1}\left(\sqrt{x-1}+2\sqrt{x+3}\right)}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}\right]=0\\ &\Leftrightarrow& \sqrt{x-1}\left(\dfrac{1}{\sqrt{x+1}}-\dfrac{2}{\sqrt{x-1}+\sqrt{x+3}}-\dfrac{2\sqrt{x+3}}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}\right)=0 \end{eqnarray*}} \item Áp dụng bất đẳng thức Bunhacopxki, ta có:
$ \sqrt{x-1}+\sqrt{x+3} \le \sqrt{(1+1)(x-1+x+3)}=\sqrt{2\cdot2(x+1)}=2\sqrt{x+1} $
\item Do đó:
$ \dfrac{2}{\sqrt{x-1}+\sqrt{x+3}}\ge \dfrac{1}{\sqrt{x+1}}\Rightarrow \dfrac{1}{\sqrt{x+1}}-\dfrac{2}{\sqrt{x-1}+\sqrt{x+3}} < 0 $.
\item Suy ra
$ \left(\dfrac{1}{\sqrt{x+1}}-\dfrac{2}{\sqrt{x-1}+\sqrt{x+3}}-\dfrac{2\sqrt{x+3}}{\left(\sqrt{x-1}+\sqrt{x+3}\right)^2}\right) < 0 $
\item Vì vậy phương trình tương đương $ \sqrt{x-1}=0 \Leftrightarrow x=1 $. Nghiệm là $ x=1 $
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