Question

Помогите решить пример
Найдите значение дроби

Answer 1

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

если $x=\frac{2ab}{b^{2}+1}$, то

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

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

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

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