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Автор - mitru

помогите срочно,как решается 4 и 5 задание​

Ответ

Автор - Helper211

Ответ:

B4) 20

B5) 2,25

Пошаговое объяснение:

В4 - прямая подстановка:

frac{a^{frac{1}{2}}-b^{frac{1}{2}}}{a^{frac{1}{4}}-b^{frac{1}{4}}}+frac{3a^{frac{1}{2}}-a^{frac{1}{4}}b^{frac{1}{4}}}{a^{frac{1}{4}}}=frac{sqrt{a} -sqrt{b} }{sqrt{sqrt{a}} -sqrt{sqrt{b}}}+frac{3sqrt{a} -sqrt{sqrt{a}} sqrt{sqrt{b}}}{sqrt{sqrt{a}}}=frac{sqrt{625} -sqrt{16} }{sqrt{sqrt{625}} -sqrt{sqrt{16}}}+frac{3sqrt{625} -sqrt{sqrt{625}} sqrt{sqrt{16}}}{sqrt{sqrt{625}}}=frac{25-4 }{sqrt{25} -sqrt{4}}+frac{3cdot 25 -sqrt{25} sqrt{4}}{sqrt{25}}=frac{21 }{5-2}+frac{75 -10}{5}=7+13=20

B5:

2^{log_2{9}+log_5{frac{1}{25}}}=2^{log_2{9}+log_5{25^{-1}}}=2^{log_2{9}-log_5{25}}=2^{log_2{9}-2}=frac{2^{log_2{9}}}{2^2}=frac{9}{4}=2,25

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