Cho x < 0 . Rút gọn biểu thức \(\sqrt{\frac{-1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}}{1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}}}\)
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Lời giải:
Báo saiTa có:
\(1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}=\frac{1}{4}\left[\left(2^{x}-2^{-x}\right)^{2}+4\right]=\frac{1}{4}\left(2^{x}+2^{-x}\right)^{2} \text { vì } 2^{x} \cdot 2^{-x}=1\)
\(\begin{array}{l} \Rightarrow \sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}=\frac{1}{2} \sqrt{\left(2^{x}+2^{-x}\right)^{2}}=\frac{1}{2}\left|2^{x}+2^{-x}\right|=\frac{1}{2}\left(2^{x}+2^{-x}\right) 2^{x}+2^{-x}>0 \\ =\left\{\begin{array}{l} -1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}=-1+\frac{1}{2}\left(2^{x}+2^{-x}\right)=\frac{1}{2}\left(2^{x}+2^{-x}-2\right)=\frac{1}{2}\left(2^{\frac{x}{2}}-2^{-\frac{x}{2}}\right)^{2} \\ 1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}=1+\frac{1}{2}\left(2^{x}+2^{-x}\right)=\frac{1}{2}\left(2^{x}+2^{-x}+2\right)=\frac{1}{2}\left(2^{\frac{x}{2}}+2^{-\frac{x}{2}}\right)^{2} \end{array}\right. \end{array}\)
\(\Rightarrow \sqrt{\frac{-1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}}{1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}}}\)
\(=\sqrt{\frac{\frac{1}{2}\left(2^{\frac{x}{2}}-2^{-\frac{x}{2}}\right)^{2}}{\frac{1}{2}\left(2^{\frac{x}{2}}+2^{-\frac{x}{2}}\right)^{2}}}\)
\(=\frac{\left|2^{\frac{x}{2}}-2^{-\frac{x}{2}}\right|}{2^{\frac{x}{2}}+2^{-\frac{x}{2}}}=\frac{\left|2^{x}-1\right|}{2^{x}+1}\)
Vì x<0 nên \(2^{x}<2^{0}=1 \Rightarrow 2^{x}-1<0 \Rightarrow\left|2^{x}-1\right|=1-2^{x}\)
Vậy \(\sqrt{\frac{-1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}}{1+\sqrt{1+\frac{1}{4}\left(2^{x}-2^{-x}\right)^{2}}}}=\frac{1-2^{x}}{1+2^{x}}\)
