é?a questão diz que posso resolver com a equação limite da soma de um pg, mas eu não consegui nem armar.
a1= x
q=
![\sqrt[]{x} \sqrt[]{x}](/latexrender/pictures/23c0d9674da78a0d1fae7f37c6ce8039.png)
S=
![\frac{x}{1-\sqrt[]{x}} \frac{x}{1-\sqrt[]{x}}](/latexrender/pictures/0df0fb8dc255d4450c8e3a985daefbc5.png)
![\frac{x}{1-\sqrt[]{x}} . \frac{1+\sqrt[]{x}}{1+\sqrt[]{x}} =\frac{x+x\sqrt[]{x}}{1-x}
}{}}{} \frac{x}{1-\sqrt[]{x}} . \frac{1+\sqrt[]{x}}{1+\sqrt[]{x}} =\frac{x+x\sqrt[]{x}}{1-x}
}{}}{}](/latexrender/pictures/46dc560c00244f48e3c781f96ee4570f.png)
é?![\sqrt[]{x} \sqrt[]{x}](/latexrender/pictures/23c0d9674da78a0d1fae7f37c6ce8039.png)
![\frac{x}{1-\sqrt[]{x}} \frac{x}{1-\sqrt[]{x}}](/latexrender/pictures/0df0fb8dc255d4450c8e3a985daefbc5.png)
![\frac{x}{1-\sqrt[]{x}} . \frac{1+\sqrt[]{x}}{1+\sqrt[]{x}} =\frac{x+x\sqrt[]{x}}{1-x}
}{}}{} \frac{x}{1-\sqrt[]{x}} . \frac{1+\sqrt[]{x}}{1+\sqrt[]{x}} =\frac{x+x\sqrt[]{x}}{1-x}
}{}}{}](/latexrender/pictures/46dc560c00244f48e3c781f96ee4570f.png)




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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}} \frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png)
(dica : igualar a expressão a
e elevar ao quadrado os dois lados)