Ответ(ы) на вопрос:
Гость
[latex] \sqrt{ \frac{1,69a^8}{b^2} } ,b\ \textless \ 0[/latex]
[latex] \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}; [/latex]
[latex]\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. [/latex]
[latex](\sqrt{ab}-\sqrt{ \frac{b}{a} }-\sqrt{ \frac{a}{b} }+\sqrt{ \frac{1}{ab} })\sqrt{ab}=(a-1)(b-1)[/latex]
[latex](\sqrt{ab}- \frac{\sqrt{b}}{\sqrt{a}}- \frac{\sqrt{a}}{\sqrt{b}}+ \frac{\sqrt{1}}{\sqrt{ab}})\sqrt{ab}=(a-1)(b-1) [/latex]
[latex]( \frac{(\sqrt{ab}^2-(\sqrt{b})^2-(\sqrt{a})^2+1)}{\sqrt{ab}} )\sqrt{ab}=(a-1)(b-1)[/latex]
[latex]( \frac{ab-b-a+1}{\sqrt{ab}})\sqrt{ab}=(a-1)(b-1) [/latex]
[latex]ab-b-a+1=(a-1)(b-1)[/latex]
[latex]ab-b-a+1=ab-a-b+1;[/latex]
[latex] \frac{2a\sqrt{ab}+a^2\sqrt{b}+2b\sqrt{ab}+ab\sqrt{b}}{2\sqrt{ab}+a\sqrt{b}}=a+b [/latex]
[latex] \frac{2\sqrt{ab}(a+b)+a\sqrt{b}(a+b)}{2\sqrt{ab}+a\sqrt{b}}=a+b [/latex]
[latex] \frac{(a+b)(2\sqrt{ab}+a\sqrt{b})}{(2\sqrt{ab}+a\sqrt{b})}=a+b [/latex]
[latex]a+b=a+b[/latex]
[latex](\sqrt{6x}-2)^2=\sqrt{3}(\sqrt{2x}-\sqrt{12})+6x[/latex]
[latex]6x-4\sqrt{6x}+4=\sqrt{6x}-6+6x[/latex]
[latex]6x-6x-4\sqrt{6x}-\sqrt{6x}=-6-4[/latex]
[latex]-5\sqrt{6x}=-10[/latex]
[latex]\sqrt{6x}=2[/latex]
[latex](\sqrt{6x})^2=2^2[/latex]
[latex]6x=4[/latex]
[latex]x= \frac{4}{6}= \frac{2}{3} [/latex]
[latex](\sqrt{7x}-2\sqrt{5})(\sqrt{7x}+2\sqrt{5})=7x-\sqrt{2}(\sqrt{5x}-\sqrt{72})[/latex]
[latex](\sqrt{7x})^2-(2\sqrt{5})^2=7x-\sqrt{10x}+\sqrt{144}[/latex]
[latex]7x-10=7x-\sqrt{10x}+12[/latex]
[latex]7x-7x+\sqrt{10x}=12+10[/latex]
[latex]\sqrt{10x}=22[/latex]
[latex](\sqrt{10x})^2=22^2[/latex]
[latex]10x=484[/latex]
[latex]x=48,4[/latex]
Не нашли ответ?
Похожие вопросы