Решите пожалуйста Номера 517, 522, 524(а,б). Created by 11032002rus. algebra-ru

Решите пожалуйста Номера 517, 522, 524(а,б)...

11032002rus @11032002rus

August 2021 1 7 Report

Решите пожалуйста

Номера 517, 522, 524(а,б)

Answers & Comments

nadelyaev2002 $\sqrt{ \frac{1,69a^8}{b^2} } ,b\ \textless \ 0$
$\frac{\sqrt{1,69a^8}}{\sqrt{b^2}}= \frac{\sqrt{13^2*(a^4)^2}}{|b|}= \frac{|13|*|a^4|}{|b|}= \frac{13a^4}{-b}= -\frac{13a^4}{b};$
$\sqrt{ \frac{0,16a^{14}}{b^{12}} }= \frac{ \sqrt{0,16a^{14}} }{ \sqrt{b^{12}} }= \frac{\sqrt{(0,4)^2*(a^7)^2}}{\sqrt{(b^6)^2}}= \frac{|0,4|*|a^7|}{|b^6|}= \frac{0,4a^7}{b^6} ,a\ \textgreater \ 0.$
$(\sqrt{ab}-\sqrt{ \frac{b}{a} }-\sqrt{ \frac{a}{b} }+\sqrt{ \frac{1}{ab} })\sqrt{ab}=(a-1)(b-1)$
$(\sqrt{ab}- \frac{\sqrt{b}}{\sqrt{a}}- \frac{\sqrt{a}}{\sqrt{b}}+ \frac{\sqrt{1}}{\sqrt{ab}})\sqrt{ab}=(a-1)(b-1)$
$( \frac{(\sqrt{ab}^2-(\sqrt{b})^2-(\sqrt{a})^2+1)}{\sqrt{ab}} )\sqrt{ab}=(a-1)(b-1)$
$( \frac{ab-b-a+1}{\sqrt{ab}})\sqrt{ab}=(a-1)(b-1)$
$ab-b-a+1=(a-1)(b-1)$
$ab-b-a+1=ab-a-b+1;$

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

$(\sqrt{6x}-2)^2=\sqrt{3}(\sqrt{2x}-\sqrt{12})+6x$
$6x-4\sqrt{6x}+4=\sqrt{6x}-6+6x$
$6x-6x-4\sqrt{6x}-\sqrt{6x}=-6-4$
$-5\sqrt{6x}=-10$
$\sqrt{6x}=2$
$(\sqrt{6x})^2=2^2$
$6x=4$
$x= \frac{4}{6}= \frac{2}{3}$

$(\sqrt{7x}-2\sqrt{5})(\sqrt{7x}+2\sqrt{5})=7x-\sqrt{2}(\sqrt{5x}-\sqrt{72})$
$(\sqrt{7x})^2-(2\sqrt{5})^2=7x-\sqrt{10x}+\sqrt{144}$
$7x-10=7x-\sqrt{10x}+12$
$7x-7x+\sqrt{10x}=12+10$
$\sqrt{10x}=22$
$(\sqrt{10x})^2=22^2$
$10x=484$
$x=48,4$

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nadelyaev2002 Если решу уравнения, обновлю ответ.

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