Giải bài 85 trang 130 SGK giải tích nâng cao 12

Ta có

$\begin{aligned} VT&=\sqrt{\dfrac{-1+\sqrt{1+\dfrac{1}{4}{{\left( {{2}^{x}}-{{2}^{-x}} \right)}^{2}}}}{1+\sqrt{1+\dfrac{1}{4}{{\left( {{2}^{x}}-{{2}^{-x}} \right)}^{2}}}}} \\ & =\sqrt{\dfrac{-1+\dfrac{\sqrt{4+{{2}^{2x}}-2+{{2}^{-2x}}}}{2}}{1+\dfrac{\sqrt{4+{{2}^{2x}}-2+{{2}^{-2x}}}}{2}}} \\ & =\sqrt{\dfrac{-1+\dfrac{\sqrt{{{\left( {{2}^{x}}+{{2}^{-x}} \right)}^{2}}}}{2}}{1+\dfrac{\sqrt{{{\left( {{2}^{x}}+{{2}^{-x}} \right)}^{2}}}}{2}}} \\ & =\sqrt{\dfrac{-1+\dfrac{{{2}^{x}}+{{2}^{-x}}}{2}}{1+\dfrac{{{2}^{x}}+{{2}^{-x}}}{2}}} \\ & =\sqrt{\dfrac{\dfrac{{{2}^{2x}}-{{2.2}^{x}}+1}{{{2}^{x}}}}{\dfrac{{{2}^{2x}}+{{2.2}^{x}}+1}{{{2}^{x}}}}} \\ & =\sqrt{\dfrac{{{\left( {{2}^{x}}-1 \right)}^{2}}}{{{\left( {{2}^{x}}+1 \right)}^{2}}}}=\dfrac{{{2}^{x}}-1}{{{2}^{x}}+1}=VP \\ \end{aligned}$

Vậy đẳng thức được chứng minh.