Question

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Answer 1

 

$\displaystyle\\\sqrt[\b{n}]{\left(x+1\right)^\b2}+\sqrt[\b{n}]{\left(x-1\right)^\b2}=4\cdot \sqrt[\b{n}]{\left(x^{\b2}-1\right)}\\\\\\\frac{\sqrt[\b{n}]{\left(x+1\right)^\b2}+\sqrt[\b{n}]{\left(x-1\right)^\b2}}{\sqrt[\b{n}]{\left(x^{\b2}-1\right)}}=4\\\\\\\\\frac{\sqrt[\b{n}]{\left(x+1\right)^\b2}}{\sqrt[\b{n}]{\left(x^{\b2}-1\right)}}+\frac{\sqrt[\b{n}]{\left(x-1\right)^\b2}}{\sqrt[\b{n}]{\left(x^{\b2}-1\right)}}=4$

$\displaystyle\\\frac{\sqrt[\b{n}]{\Big(x+1\Big)\Big(x+1\Big)}}{\sqrt[\b{n}]{\Big(x+1\Big)\Big(x-1\Big)}}+\frac{\sqrt[\b{n}]{\Big(x-1\Big)\Big(x-1\Big)}}{\sqrt[\b{n}]{\Big(x+1\Big)\Big(x-1\Big)}}=4\\\\\\\\\sqrt[\b{n}]{\frac{\Big(x+1\Big)\Big(x+1\Big)}{\Big(x+1\Big)\Big(x-1\Big)}}+ \sqrt[\b{n}]{\frac{\Big(x-1\Big)\Big(x-1\Big)}{\Big(x+1\Big)\Big(x-1\Big)}}=4~~~~~\text{(Simplificam)}$

$\displaystyle\\\sqrt[\b{n}]{\frac{x+1}{x-1}}+\sqrt[\b{n}]{\frac{x-1}{x+1}}=4\\\\\\\sqrt[\b{n}]{\frac{x+1}{x-1}}+\frac{1}{\sqrt[\b{n}]{\dfrac{x+1}{x-1}}}=4\\\\\\ \text{Substitutie: }~~~\boxed{y=\sqrt[\b{n}]{\frac{x+1}{x-1}}}\\\\\\y+\frac{1}{y}=4\\\\\\y-4+\frac{1}{y}=0~\Big|\cdot y\\\\\\y^2-4y+1=0\\\\y_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-4}}{2} =\frac{4\pm\sqrt{12}}{2}=\frac{4\pm2\sqrt{3}}{2}=2\pm\sqrt{3}\\\\\boxed{y_1=2-\sqrt{3}}\\\\\boxed{y_2=2+\sqrt{3}}$

$\displaystyle\\\text{Revenim la substitutie:}\\\\y=\sqrt[\b{n}]{\frac{x+1}{x-1}}\\\\y_1=2-\sqrt{3}=\sqrt[\b{n}]{\frac{x+1}{x-1}}\\\\\frac{x+1}{x-1}=\Big(2-\sqrt{3}\Big)^n\\\\(x-1)\Big(2-\sqrt{3}\Big)^n=x+1\\\\\Big(2-\sqrt{3}\Big)^nx-\Big(2-\sqrt{3}\Big)^n-x=1\\\\x\left(\Big(2-\sqrt{3}\Big)^n-1\right)=\Big(2-\sqrt{3}\Big)^n+1\\\\\\\boxed{x_1=\frac{\Big(2-\sqrt{3}\Big)^n+1}{\Big(2-\sqrt{3}\Big)^n-1}}$

$\displaystyle\\\text{Prin analogie }x_2\text{ este:}\\\\\\\boxed{x_2=\frac{\Big(2+\sqrt{3}\Big)^n+1}{\Big(2+\sqrt{3}\Big)^n-1}}$