Решите пж срочно надо, умоляю:

Question 1

Question 2

№ 7
1.
$x \leq 0; y \geq 0 \\ \\ \sqrt{25y} = 5 \sqrt{y} \\\sqrt{0,01x^2} =0,1|x|=-0,1x \\ \sqrt{2,89x^6y^5} = 1,7|x^3|y^2 \sqrt{y} =-1,7x^3y^2 \sqrt{y}$

2.
$5 \sqrt{2} = \sqrt{50} \\ 7x \sqrt{xy} = \sqrt{49x^3y} \\ \frac{6a \sqrt{x} }{ \sqrt{3a} } = \sqrt{ \frac{36a^2x}{3a} } = \sqrt{12ax}$

3.
$\sqrt{8} - \sqrt{18} + \sqrt{3} =2 \sqrt{2} -3 \sqrt{2} + \sqrt{3} = - \sqrt{2} + \sqrt{3} \\ \\ \sqrt{20} -2 \sqrt{2}( \sqrt{10} -4 \sqrt{2} )=2 \sqrt{5} -4\sqrt{5} +16=-2 \sqrt{5} +16$

4.
$\frac{7}{2 \sqrt{7} } = \frac{ \sqrt{7} }{2} \\ \\ \frac{2x+ \sqrt{x} }{4 \sqrt{x} } = \frac{ \sqrt{x} (2 \sqrt{x} +1)}{4 \sqrt{x} } = \frac{2 \sqrt{x} +1}{4} \\ \\ \frac{25 - x}{5+ \sqrt{x} } = \frac{(5- \sqrt{x} )(5+ \sqrt{x} )}{5+ \sqrt{x} } =5- \sqrt{x} \\ \\ \frac{27-a \sqrt{a} }{3- \sqrt{a} } = \frac{(3- \sqrt{a} )(a+3 \sqrt{a} +9)}{3- \sqrt{a} } = a+3 \sqrt{a}+9$

5.
$\frac{2}{ \sqrt{6} } = \frac{2 \sqrt{6} }{6} = \frac{\sqrt{6} }{3} \\ \\ \frac{1}{3 \sqrt{7} } = \frac{ \sqrt{7} }{21}$

$\frac{4}{ \sqrt{7}- \sqrt{3} } = \frac{4( \sqrt{7}+ \sqrt{3} )}{( \sqrt{7} - \sqrt{3} )( \sqrt{7} + \sqrt{3} )} = \frac{4( \sqrt{7} + \sqrt{3}) }{7-3} = \sqrt{7} + \sqrt{3} \\ \\ \frac{11}{5- \sqrt{14} } - \frac{2}{4- \sqrt{14} } = \frac{11(5+ \sqrt{14} )}{25-14} - \frac{2(4+ \sqrt{14} )}{16-14} =5+ \sqrt{14} -4 - \sqrt{14} =1$

6.
$\frac{ \sqrt{a^3}- \sqrt{b^3} }{a-b} - \frac{ \sqrt{a^3}+ \sqrt{b^3} }{a+b}= \frac{( \sqrt{a^3}- \sqrt{b^3})(a+b)-( \sqrt{a^3}+ \sqrt{b^3})(a-b)}{(a-b)(a+b)} = \\ \\ = \frac{a^2 \sqrt{a}-ab \sqrt{b} +ab \sqrt{a} -b^2 \sqrt{b} -a^2 \sqrt{a} -ab \sqrt{b} +ab \sqrt{a} +b^2 \sqrt{b} }{(a-b)(a+b)} = \\ \\ = \frac{2ab \sqrt{a} -2ab \sqrt{b} }{(a-b)(a+b)} = \frac{2ab( \sqrt{a}- \sqrt{b} )}{(a-b)(a+b)}$

$a= \sqrt{3} - \sqrt{2} ;b= \sqrt{3} + \sqrt{2} \\ \\ \frac{2ab( \sqrt{a}- \sqrt{b} )}{(a-b)(a+b)}= \frac{2( \sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})( \sqrt{\sqrt{3} - \sqrt{2}}- \sqrt{\sqrt{3} + \sqrt{2}} )}{(\sqrt{3} - \sqrt{2}- \sqrt{3}- \sqrt{2} )(\sqrt{3} - \sqrt{2}+ \sqrt{3} + \sqrt{2} )}= \\ \\ = \frac{2*1*( \sqrt{\sqrt{3} - \sqrt{2}}-\sqrt{\sqrt{3} + \sqrt{2}})}{-2 \sqrt{2}*2 \sqrt{3} } =- \frac{ \sqrt{\sqrt{3} - \sqrt{2}}-\sqrt{\sqrt{3} + \sqrt{2}}}{2 \sqrt{2} \sqrt{3} } =$
$=- \frac{ \sqrt{\sqrt{3} - \sqrt{2}}}{2 \sqrt{2} \sqrt{3} }+ \frac{ \sqrt{\sqrt{3} +\sqrt{2}}}{2 \sqrt{2} \sqrt{3} } =- \frac{1}{2} \sqrt{ \frac{ \sqrt{3} - \sqrt{2} }{6 } } + \frac{1}{2} \sqrt{ \frac{ \sqrt{3} + \sqrt{2} }{6} }$

7.
$\sqrt{10} \\ 2 \sqrt{3} = \sqrt{12} \\ 3= \sqrt{9} 3; \sqrt{10} ;2 \sqrt{3}$

8.
$( \frac{x \sqrt{x} -y \sqrt{y} }{ \sqrt{x} - \sqrt{y} } + \sqrt{xy} ):( \frac{x-y}{ \sqrt{x} - \sqrt{y} } )^2= \\ \\ =(\frac{(x \sqrt{x} -y \sqrt{y} )( \sqrt{x} + \sqrt{y} )}{( \sqrt{x} - \sqrt{y})( \sqrt{x} + \sqrt{y} ) } + \sqrt{xy} ):( \frac{ (\sqrt{x} - \sqrt{y}) ( \sqrt{x} + \sqrt{y} )}{ \sqrt{x} - \sqrt{y} } )^2=$
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