Помогите, плиз! Заранее спасибо.

Question 1

Помогите, плиз!
Заранее спасибо.

Question 2

Решите задачу:

$1) (1+3 \sqrt{5} ) ^{2} =1+45+6 \sqrt{5} =46+6 \sqrt{5} \\ 2)(2 \sqrt{3} -7) ^{2} =12+49-28 \sqrt{2} = \\ =61-28 \sqrt{2} \\ 3)(2 \sqrt{10} - \sqrt{2} ) ^{2} =40+4-4 \sqrt{20} = \\ =44-4*2 \sqrt{5} =44-8 \sqrt{5} \\ \\ 1) \frac{ \sqrt{x} -5}{25-x} =- \frac{5- \sqrt{x} }{25-x} =- \frac{5- \sqrt{x} }{(5- \sqrt{x} )(5+ \sqrt{x} )} = \\ =- \frac{1}{5+ \sqrt{x} } \\ \\ 2) \frac{ \sqrt{2} +2}{ \sqrt{2} } = \frac{ \sqrt{2} (1+ \sqrt{2} )}{ \sqrt{2} } =1+ \sqrt{2} \\ \\$

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

Question 3

1) (1+3√5)²=1²+2*1*3√5+(3√5)²=1+6√5+45=6√5+46
2)(2√3-7)²=(2√3)²-2*2√3*7+7²=12-28√3+49=61-28√3
3)(2√10-√2)²=(2√10)²-2*2√10*√2+(√2)²=40-4√20+2=42-4*2√5=42-8√5

1) $\frac{ \sqrt{x} -5}{25-x} =- \frac{ \sqrt{x} -5}{x-25} =- \frac{ \sqrt{x} -5}{( \sqrt{x} )^{2} - 5^{2} } =- \frac{ \sqrt{x} -5}{( \sqrt{x} - 5)( \sqrt{x} + 5)} } =- \frac{1}{ \sqrt{x} +5}$
2) $\frac{ \sqrt{2}+2 }{ \sqrt{2} }= \frac{ \sqrt{2}( \sqrt{2}+2 )}{ \sqrt{2}* \sqrt{2} } \frac{2+2 \sqrt{2}}{2} = \frac{2(1+ \sqrt{2})}{2} =1+ \sqrt{2}$
3) $\frac{5+ \sqrt{10} }{\sqrt{10}} = \frac{\sqrt{10}(5+\sqrt{10})}{\sqrt{10}*\sqrt{10}} = \frac{5\sqrt{10}+10}{10} = \frac{5(\sqrt{10}+2)}{10} = \frac{\sqrt{10}+2}{2}$
4) $\frac{2 \sqrt{3}-3 }{5\sqrt{3}} = \frac{\sqrt{3}(2\sqrt{3}-3)}{5\sqrt{3}*\sqrt{3}} = \frac{6-3\sqrt{3}}{5*3} = \frac{3(2-\sqrt{3})}{5*3} = \frac{2-\sqrt{3}}{5}$