Решите, с обьяснениями. Created by belajasnezhinka. matematika-ru

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Решите, с обьяснениями...

Amway96rus 1) $7 \sqrt{27a}-2 \sqrt{75a} + \sqrt{300a}=7 \sqrt{9*3a}-2 \sqrt{25*3a}+ \sqrt{100*3a}= \\ 
21 \sqrt{3a}-10 \sqrt{3a}+10 \sqrt{3a}=21 \sqrt{3a} \\$
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2) $(2 \sqrt{3}-5 \sqrt{11} )^2=12-2*2 \sqrt{3}*5 \sqrt{11}+275=287-20 \sqrt{33}$
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3) (13-25a²)=(√(13)-5a)(√13+5a)
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4) $\frac{ \sqrt{15}- \sqrt{10}+5}{ \sqrt{3}- \sqrt{2}+ \sqrt{5} }= \frac{ \sqrt{5}( \sqrt{3}- \sqrt{2})+5 }{ \sqrt{3}- \sqrt{2}+ \sqrt{5} }$
пусть √3-√2 = x, тогда
$\frac{ \sqrt{5}x+5 }{x+ \sqrt{5} } = \frac{ \sqrt{5}( x+ \sqrt{5}) }{x+ \sqrt{5} } = \sqrt{5}$
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5) $\frac{ \sqrt{x} }{ \sqrt{x} -3 \sqrt{y} }= \frac{ \sqrt{x} ( \sqrt{x} +3 \sqrt{y}) }{ (\sqrt{x} -3 \sqrt{y})( \sqrt{x} +3 \sqrt{y}) }= \frac{x+3 \sqrt{xy} }{x-9y}$

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