Упростить выражения См. фото

Question 1

Question 2

1.036 не надо

Question 3

1.059.

$\sqrt{ \frac{ \sqrt{2} }{a}+ \frac{a}{ \sqrt{2} } +2 }= \sqrt{ \frac{( \sqrt{2})^2 +a^2+2a \sqrt{2} }{a \sqrt{2} } }= \sqrt{ \frac{(a+ \sqrt{2})^2 }{a \sqrt{2} } }= \frac{a+ \sqrt{2} }{ \sqrt{a\cdot \sqrt[4]{2} } }$

$a \sqrt{2a}- \sqrt[4]{8a^4}= a \sqrt{2a}-\sqrt[4]{(2 \sqrt{2} a^2)^2}=a \sqrt{2a}-\sqrt{2 \sqrt{2} a^2}= \\ \\ =a \sqrt{2}\cdot( \sqrt{a} - \sqrt[4]{2})$

$\frac{a+ \sqrt{2} }{ \sqrt{a}\cdot \sqrt[4]{2} } - \frac{a^2 \sqrt[4]{2} -2 \sqrt{a} }{a \sqrt{2}\cdot( \sqrt{a}- \sqrt[4]{2}) } = \frac{(a +\sqrt{2})( \sqrt{a}- \sqrt[4]{2})\cdot \sqrt{a}\cdot \sqrt[4]{2} -a^2 \sqrt[4]{2}+2 \sqrt{a}}{a \sqrt{2}\cdot( \sqrt{a}- \sqrt[4]{2})}=$

$= \frac{a^2 \sqrt[4]{2}-a \sqrt{a}\cdot \sqrt{2}+a \sqrt{2}\cdot \sqrt[4]{2}- \sqrt{2}\cdot \sqrt{a}\cdot \sqrt{2}-a^2 \sqrt{2}+2 \sqrt{a} }{a \sqrt{2}\cdot( \sqrt{a}- \sqrt[4]{2})}= \\ \\ = \frac{-a \sqrt{a}\cdot \sqrt{2}+a \sqrt{2}\cdot \sqrt[4]{2}}{a \sqrt{2}\cdot( \sqrt{a}- \sqrt[4]{2})}= \frac{ \sqrt{a}\cdot \sqrt{2}( \sqrt[4]{2}- \sqrt{a})}{a \sqrt{2}\cdot( \sqrt{a} - \sqrt[4]{2}) } =-1$

1.058.

$\frac{x\cdot \frac{1}{ \sqrt{x^2-a^2} +1} }{ \frac{a}{ \sqrt{x-a} }+ \sqrt{x-a} }= \frac{x+ \sqrt{x^2-a^2} }{ \sqrt{x^2-a^2} } : \frac{a+(x-a)}{ \sqrt{x-a} } =\frac{x+ \sqrt{x^2-a^2} }{ \sqrt{(x-a)(x+a)} } \cdot \frac{ \sqrt{x-a} }{x}=\frac{x+ \sqrt{x^2-a^2} }{ x\sqrt{x+a} }$

$\frac{x+ \sqrt{x^2-a^2} }{ x\sqrt{x+a} } :\frac{a^2 \sqrt{x+a}} {x- \sqrt{x^2-a^2} } =\frac{x+ \sqrt{x^2-a^2} }{ x\sqrt{x+a} } \cdot \frac{x- \sqrt{x^2-a^2} }{a^2 \sqrt{x+a} }= \frac{x^2-(x^2-a^2)}{a^2x(x+a)}= \frac{1}{x(x+a)}$

$\frac{1}{x(x+a)} + \frac{1}{x^2-ax}= \frac{1}{x(x+a)} + \frac{1}{x(x-a)}= \\ \\ = \frac{x-a+x+a}{x(x+a)(x-a)}= \frac{2}{(x+a)(x-a)}$