Question

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Даю 30 балівв ​

Answer 1

Ответ:

Пошаговое объяснение:

$\frac{\sqrt{a} +\sqrt{b} }{a-b} -\frac{1}{\sqrt{a} -\sqrt{b}} =\frac{\sqrt{a} +\sqrt{b} }{a-b} -\frac{\sqrt{a} +\sqrt{b}}{(\sqrt{a} -\sqrt{b})(\sqrt{a} +\sqrt{b})} =\frac{\sqrt{a} +\sqrt{b} }{a-b} -\frac{\sqrt{a} +\sqrt{b} }{a-b} =0$

$\frac{a^3+b^3}{a+b} +3ab=\frac{(a+b)(a^2-ab+b^2)}{a+b}+3ab=a^2-ab+b^2+3ab=a^2+2ab+b^2 = (a+b)^2$

$4a^2+4a=4a(a+1)$

Answer 2

Ответ:

$2)\ \ \displaystyle \frac{\sqrt{a}+\sqrt{b}}{a-b}-\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a}+\sqrt{b}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}-\frac{1}{\sqrt{a}-\sqrt{b}}=\\\\\\=\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{1}{\sqrt{a}-\sqrt{b}}=0$

$3)\ \ \displaystyle \frac{a^3+b^3}{a+b}+3ab=\frac{(a+b)(a^2-ab+b^2)}{a+b}+3ab=\\\\\\=a^2-ab+b^2+3ab=a^2+2ab+b^2=(a+b)^2\\\\\\4)\ \ 4a^2+4a=4a\cdot (a+1)$