Question

$\rm Resolva,em~\mathbb{R},a~equac_{\!\!,}\tilde ao~\dfrac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\dfrac{11}{5}.$

Answer 1

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$\boxed{\begin{array}{l}\sf \dfrac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\dfrac{11}{5}\\\sf 11\cdot(x+\sqrt{x+1})=5\cdot(x-\sqrt{x+1})\\\sf11x+11\sqrt{x+1}=5x-5\sqrt{x+1}\\\sf 11\sqrt{x+1}+5\sqrt{x+1}=5x-11x\\\sf16\sqrt{x+1}=-6x\div(2)\\\sf 8\sqrt{x+1}=-3x\\\sf(8\sqrt{x+1})^2=(-3x)^2\\\sf 64\cdot(x+1)=9x^2\\\sf 9x^2=64x+64\\\sf 9x^2-64x-64=0\\\sf\Delta=4096+2304=6400\\\sf x=\dfrac{64\pm80}{18}\begin{cases}\sf x_1=8\\\sf x_2=-\dfrac{8}{9}\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm verificac_{\!\!,}\tilde ao\!:}\\\sf para~x=8:\\\sf\dfrac{8-\sqrt{8+1}}{8+\sqrt{8+1}}=\dfrac{8-3}{8+3}=\dfrac{5}{11}\ne\dfrac{11}{5}.\\\sf para~x=-\dfrac{8}{9}:\\\sf \dfrac{-\frac{8}{9}-\sqrt{-\frac{8}{9}+1}}{-\frac{8}{9}+\sqrt{-\frac{8}{9}+1}}=\dfrac{-\frac{8}{9}-\sqrt{\frac{1}{9}}}{-\frac{8}{9}+\sqrt{\frac{1}{9}}}\\\sf=\dfrac{-\frac{8}{9}-\frac{1}{3}}{-\frac{8}{9}+\frac{1}{3}}=\dfrac{-\frac{11}{\diagup\!\!\!9}}{-\frac{5}{\diagup\!\!\!9}}=\dfrac{11}{5}\end{array}}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf S=\bigg\{-\dfrac{8}{9}\bigg\}}}}}$