Question

срочно прошу помогите

сократите дробь 1) 5-√5/√10-5√2

4а^2+4а√б+б/4а^2-б

3) √3-3/3√2-√6

4) 9а^2-б^2/9а-6б√а+б^2

освободитесь от знака корня в знаменателе

1)10/3√5

2)11/2√3+1

3) 15/2√6

4)19/2√5-1

Answer 1

$1) frac{5-sqrt{5}}{sqrt{10}-5sqrt{2}}=frac{5-sqrt{5}}{sqrt{10}-sqrt{50}}=frac{(sqrt{10}-sqrt{50})(5-sqrt{5})}{(sqrt{10}+sqrt{50})(sqrt{10}-sqrt{50})}=frac{(sqrt{10}-sqrt{50})(5-sqrt{5})}{10-50}=-frac{10sqrt{10}-30sqrt{2}}{40}\=0,75sqrt{2}-0,25sqrt{10}$

 

$2)frac{ 4a^2+4asqrt{b}+b}{4a^2-b}=frac{(2a+sqrt{b})^2}{(2a+sqrt{b})(2a-sqrt{b})}=frac{2a+sqrt{b}}{2a-sqrt{b}}$

 

$3) frac{sqrt{3}-3}{3sqrt{2}-sqrt{6}}= frac{sqrt{3}-3}{3sqrt{2}-sqrt{2}sqrt{3}}= frac{sqrt{3}-3}{-sqrt{2}(sqrt{3}-3)}=-frac{1}{sqrt{2}}=-frac{sqrt{2}}{2}$

 

Может опечатка ,а должно быть так:

$4)frac{ 9a-b^2}{9a-6bsqrt{a}+b^2}=frac{ (3sqrt{a}-b)(3sqrt{a}+b)}{(3sqrt{a}-b)^2}=frac{ 3sqrt{a}+b}{3sqrt{a}-b}$

 

 

$1)frac{10}{3sqrt{5}}=frac{10*sqrt{5}}{3sqrt{5}*sqrt{5}}=frac{10*sqrt{5}}{15}$

 

$2)frac{11}{2sqrt{3}+1}=frac{11(2sqrt{3}-1)}{(2sqrt{3}+1)(2sqrt{3}-1)}=frac{22sqrt{3}-11}{(2sqrt{3})^2-1^2}=frac{22sqrt{3}-11}{11}=2sqrt{3}-1$

 

$3)frac{15}{2sqrt{6}}=frac{15*sqrt{6}}{2sqrt{6}*sqrt{6}}=frac{15*sqrt{6}}{2*6}=frac{15*sqrt{6}}{12}$

 

$4)frac{19}{2sqrt{5}-1}=frac{19*(2sqrt{5}+1)}{(2sqrt{5}-1)*(2sqrt{5}+1)}=frac{38sqrt{5}+19}{(2sqrt{5})^2-1^2}=frac{38sqrt{5}+19}{19}=2sqrt{5}+1$