Question

4.25 выражение.
3) а+2√а+1/а-1 + 2√а/√а-1.
6)( √а+1/√а-1 - √а-1/√а+1 + 4√а) * ( √а/4 - 1/√4а).

Answer 1

$1)\; \; \frac{a+2\sqrt{a}+1}{a-1}+\frac{2\sqrt{a}}{\sqrt{a}-1}=\frac{(\sqrt{a}+1)^2}{(\sqrt{a}-1)(\sqrt{a}+1)}+\frac{2\sqrt{a}}{\sqrt{a}-1}=\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{2\sqrt{a}}{\sqrt{a}-1}=\frac{3\sqrt{a}+1}{\sqrt{a}-1}$

$2)\; \; \Big (\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\Big )\cdot \Big (\frac{\sqrt{a}}{4}-\frac{1}{\sqrt{4a}}\Big )=\\\\=\Big (\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2}{(\sqrt{a}-1)(\sqrt{a}+1)}+4\sqrt{a}\Big )\cdot \frac{2a-4}{4\sqrt{4a}}=\\\\=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}(a-1)}{(\sqrt{a}-1)(\sqrt{a}+1)}+\frac{2(a-2)}{8\sqrt{a}}=\\\\=\frac{4\sqrt{a}+4a\sqrt{a}-4\sqrt{a}}{(\sqrt{a}-1)(\sqrt{a}+1)}+\frac{a-2}{4\sqrt{a}}=\frac{4a\sqrt{a}}{(\sqrt{a}-1)(\sqrt{a}+1)}+\frac{a-2}{4\sqrt{a}}=\frac{16a^2+(a-2)(a-1)}{4\sqrt{a}(a-1)}=\\\\=\frac{16a^2+a^2-3a+2}{4\sqrt{a}(a-1)}=\frac{17a^2-3a+2}{4\sqrt{a}(a-1)}$

Answer 2

Объяснение:

(а+2√(а+1))/(а-1) +(2√а)/(√а-1)=(а+2√(а+1))/((√a -1)(√a +1)) +(2√а)/(√а-1)=(а+2√(а+1)+2√a ·(√a +1))/(a-1)=(a+√(4(a+1))+2a+2√a)/(a-1)=(3a+√(4a+4)+√4a)/(a-1)