Question

Сократите дробь:

1) $frac{( sqrt{3}- sqrt{2})* sqrt{72} }{3(2 sqrt{6}- sqrt{16})( sqrt[3]{64}+1) }$

2)$frac{ sqrt{6}+sqrt{3}- sqrt{2}-1 }{ sqrt{6}+2 sqrt{3}- sqrt{2}-2 }$

помогите пожалуйста.срочно надо.

Answer 1

1)
$frac{(sqrt{3}-sqrt{2})*sqrt{72}}{3(2sqrt{6}-sqrt{16})(sqrt[3]{64}+1)}=\\frac{(sqrt{3}-sqrt{2})*sqrt{2}*6}{3*sqrt{2}*2(sqrt{3}-sqrt{2})*(4+1)}=\\frac{2}{5}=0.4$
так как
$sqrt{72}=sqrt{2*36}=sqrt{2}*sqrt{36}=sqrt{2}*6$
$sqrt[3]{64}=sqrt[3]{4^3}=4$
;
$2sqrt{6}-sqrt{16}=2sqrt{3}*sqrt{2}-sqrt{4}*sqrt{2}*sqrt{2}=\\2*sqrt{2}*(sqrt{3}-sqrt{2})$
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2)
$frac{sqrt{6}+sqrt{3}-sqrt{2}-1}{sqrt{6}+2sqrt{3}-sqrt{2}-2}=\\frac{sqrt{3}*sqrt{2}+sqrt{3}*1-sqrt{2}*1-1*1}{sqrt{3}*sqrt{2}+sqrt{3}*2-1*sqrt{2}-1*2}=\\frac{sqrt{3}(sqrt{2}+1)-1*(sqrt{2}+1)}{sqrt{3}(sqrt{2}+2)-1(sqrt{2}+2)}=\\frac{(sqrt{3}-1)(sqrt{2}+1)}{(sqrt{3}-1)(sqrt{2}+2)}=frac{sqrt{2}+1}{sqrt{2}+2}$