Если (20a)/(a+1)=2^1/2 тогда найти (14a)/(a^2-1)=??? a>1

Решите задачу:

$\frac{20a}{a+1}=2^{1/2}=\sqrt2\; \; ,\; \; a\ \textgreater \ 1\\\\1)\; \; \frac{14a}{a^2-1}=\frac{14a}{(a-1)(a+1)}=\frac{20a}{a+1}\cdot \frac{\frac{14}{20}}{a-1}= \sqrt2\cdot \frac{14}{20(a-1)}= \frac{7\sqrt2}{10(a-1)}\\\\\\2)\; \; \Big ( \frac{20a}{a+1} \Big )^2=(\sqrt2)^2\quad \Rightarrow \quad \Big (\frac{20a}{a+1}\Big )^2-(\sqrt2)^2=0\\\\\Big ( \frac{20a}{a+1} -\sqrt2\Big )\cdot \Big ( \frac{20a}{a+1} +\sqrt2\Big )=0\\\\a)\; \; \frac{20a}{a+1}-\sqrt2=0\; \; ,\; \; \frac{20a}{a+1} =\sqrt2\; \; ,\; \; 20a=\sqrt2a+\sqrt2$

$(20-\sqrt2)a=\sqrt2\quad \Rightarrow \; \; \; a= \frac{\sqrt2}{20-\sqrt2} \approx 0,076\ \textless \ 1\\\\b)\; \; \frac{20a}{a+1}+\sqrt2=0\; \; ,\; \; \frac{20a}{a+1}=-\sqrt2 \; \; ,\; \; 20a=-\sqrt2a-\sqrt2\\\\(20+\sqrt2)a=-\sqrt2\; \; \; \Rightarrow \; \; \; a=-\frac{\sqrt2}{20+\sqrt2}\approx -0,066\ \textless \ 1\\\\Otvet:\; \; pri\; \; a\ \textgreater \ 1\; \; net\; reshenij\; .$