Question

помогите пожалуйста ​

Answer 1

$1)\; \; a\geq -3\; \; \to \; \; a+3\geq 0\\\\a-3+\sqrt{6a+\sqrt{a^4+18a^2+81}}=a-3+\sqrt{6a+\sqrt{(a^2+9)^2}}=\\\\=a-3+\sqrt{6a+a^2+9}=a-3+\sqrt{(a+3)^2}=a-3+|\underbrace {a+3}_{\geq 0}|=\\\\=a-3+(a+3)=2a\\\\\\2)\; \; x\geq -5\; \; \to \; \; x+5\geq 0\\\\2x-1+\sqrt{10x+22+\sqrt{x^4+6x^2+9}}=2x-1+\sqrt{10x+22+\sqrt{(x^2+3)^2}}=\\\\=2x-1+\sqrt{10x+22+(x^2+3)}=2x-1+\sqrt{x^2+10x+25}=\\\\=2x-1+\sqrt{(x+5)^2}=2x-1+|\underbrace {x+5}_{\geq 0}|=2x-1+(x+5)=3x+4$

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

$4)\; \; \frac{a^2+a\sqrt2}{a^2+2}\cdot \Big (\frac{a}{a-\sqrt2}-\frac{\sqrt2}{a+\sqrt2}\Big )=\frac{a(a+\sqrt2)}{a^2+2}\cdot \frac{a(a+\sqrt2)-\sqrt2(a-\sqrt2)}{(a-\sqrt2)(a+\sqrt2)}=\\\\=\frac{a(a+\sqrt2)}{a^2+2}\cdot \frac{a^2+a\sqrt2-a\sqrt2+2}{(a-\sqrt2)(a+\sqrt2)}=\frac{a(a^2+2)}{(a^2+2)(a-\sqrt2)}=\frac{a}{a-\sqrt2}$