Question

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Answer 1

$frac{sqrt{a+x}+sqrt{a-x}}{sqrt{a+x}-sqrt{a-x}}=frac{(sqrt{a+x}+sqrt{a-x})(sqrt{a+x}+sqrt{a-x})}{(sqrt{a+x}-sqrt{a-x})(sqrt{a+x}+sqrt{a-x})}=frac{(sqrt{a+x}+sqrt{a-x})^{2}}{a+x-(a-x)}=frac{a+x+2sqrt{(a+x)(a-x)}+a-x}{a+x-a+x}=frac{a+2sqrt{a^{2}-x^{2}}+a}{2x}=frac{2a+2sqrt{a^{2}-x^{2}}}{2x}=frac{2(a+sqrt{a^{2}-x^{2}})}{2x}=frac{a+sqrt{a^{2}-x^{2}}}{x}$

если $x=frac{2ab}{b^{2}+1}$, то

$frac{a+sqrt{a^{2}-(frac{2ab}{b^{2}+1})^{2}}}{frac{2ab}{b^{2}+1}}=frac{a+sqrt{a^{2}-frac{4a^{2}b^{2}}{(b^{2}+1)^{2}}}}{frac{2ab}{b^{2}+1}}=frac{a+sqrt{frac{(b^{2}+1)^{2}a^{2}-4a^{2}b^{2}}{(b^{2}+1)^{2}}}}{frac{2ab}{b^{2}+1}}=frac{a+sqrt{frac{((b^{2}+1)a)^{2}-4a^{2}b^{2}}{(b^{2}+1)^{2}}}}{frac{2ab}{b^{2}+1}}=$

$frac{a+sqrt{frac{(ab^{2}+a)^{2}-4a^{2}b^{2}}{(b^{2}+1)^{2}}}}{frac{2ab}{b^{2}+1}}=frac{(a+sqrt{frac{(ab^{2}+a)^{2}-4a^{2}b^{2}}{(b^{2}+1)^{2}}})times (b^{2}+1)}{2ab}=frac{(a+sqrt{frac{a^{2}b^{4}+2a^{2}b^{2}+a^{2}-4a^{2}b^{2}}{(b^{2}+1)^{2}}})times (b^{2}+1)}{2ab}=$

$frac{(a+sqrt{frac{a^{2}b^{4}-2a^{2}b^{2}+a^{2}}{(b^{2}+1)^{2}}})times (b^{2}+1)}{2ab}=frac{(a+frac{sqrt{a^{2}b^{4}-2a^{2}b^{2}+a^{2}}}{b^{2}+1})times (b^{2}+1)}{2ab}=frac{(a+frac{sqrt{(ab^{2}-a)^{2}}}{b^{2}+1})times (b^{2}+1)}{2ab}=$

$frac{(a+frac{ab^{2}-a}{b^{2}+1})times (b^{2}+1)}{2ab}=frac{frac{atimes(b^{2}+1)+ab^{2}-a}{b^{2}+1}times (b^{2}+1)}{2ab}=frac{ab^{2}+a+ab^{2}-a}{2ab}=frac{2ab^{2}}{2ab}=b$

Answer 2

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