Question

Допоможіть будь ласка

Answer 1

Ответ:

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Answer 2

Объяснение:

1.

$a)\ (\sqrt{20}-2\sqrt{45}-\sqrt{125})*\sqrt{10}+5\sqrt{2}=(\sqrt{4*5} -2\sqrt{9*5}-\sqrt{25*5})*\sqrt{10}+5\sqrt{2}=\\=(2\sqrt{5}-6\sqrt{5} -5\sqrt{5})*\sqrt{10} +5\sqrt{2}=-9\sqrt{5}*\sqrt{10} +2\sqrt{5}=-9\sqrt{5*10} +5\sqrt{2}=\\=-9*\sqrt{25*2}+5\sqrt{2}=-45\sqrt{2}+5\sqrt{2} =-40\sqrt{2}.\\b)\ \frac{x^2-11}{x+\sqrt{11} } =\frac{(x+\sqrt{11})*(x-\sqrt{11}) }{x+\sqrt{11} } =x-\sqrt{11}.$

$c)\ \frac{(\sqrt{3}+1)^2 }{2+\sqrt{3} } =\frac{(\sqrt{3})^2+2\sqrt{3}*1+1^2 }{2+\sqrt{3} }=\frac{3+2\sqrt{3} +1}{2+\sqrt{3} } =\frac{4+2\sqrt{3} }{2+\sqrt{3} } =\frac{2*(2+\sqrt{3} )}{2+\sqrt{3} }=2.$

2.

$a)\ \frac{2a}{\sqrt{2ax} }=\frac{2a*\sqrt{2ax}}{\sqrt{2ax}*\sqrt{2ax}}=\frac{2a*\sqrt{2ax}}{2ax}=\frac{\sqrt{2ax}}{x}.\\b)\ \frac{1}{2-\sqrt{3} }=\frac{2+\sqrt{3} }{(2-\sqrt{3})*(2+\sqrt{3)} } =\frac{2+\sqrt{3} }{2^2-(\sqrt{3})^2 } =\frac{2+\sqrt{3} }{4-3}=\frac{2+\sqrt{3} }{1} =2+\sqrt{3}.\\c)\ \frac{a-1}{\sqrt{a+3}-2 }= \frac{(a-1)*( \sqrt{a+3}+2) }{(\sqrt{a+3}-2)*(\sqrt{a+3}+2) } =\frac{(a-1)*(\sqrt{a+3}+2)}{(\sqrt{a+3})^2 -2^2}=\frac{(a-1)*(\sqrt{a+3}+2)}{a+3-4}=$

$=\frac{(a-1)*(\sqrt{a+3}+2)}{a-1}=\sqrt{a+3}+2.$