Hihi =))) Cơ mà giỏi thật sự luôn. Mẹ tớ khen suốt thôi í 

Người ta đẹp trong con mắt kẻ si tình mà Thơ ui =)))  

Với mọi x ta có :

\(\left|x+5\right|\ge0\)

\(\Leftrightarrow\left|x+5\right|+5\ge0\)

\(\Leftrightarrow A\ge5\)

Dấu "=" xảy ra \(\Leftrightarrow x=-5\)

Vậy..

\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

\(\dfrac{2}{20^{10}-1}>\dfrac{2}{20^{10}-3}\Leftrightarrow A>B\)

ĐKXĐ : \(0\le x\le1\)

\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)^2.\dfrac{x^2-1}{2}-\sqrt{1-x^2}\)

\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)^2.\dfrac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{\left(1-x\right)\left(1+x\right)}\)

\(=\dfrac{4x}{\left(x-1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{\left(1-x\right)\left(1+x\right)}\)

\(=\dfrac{2x}{x-1}-\sqrt{1-x}.\sqrt{1+x}\)

\(=\dfrac{2x-\sqrt{1-x^2}.\left(x-1\right)}{x-1}\)

a/ \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{2}\)

b/ \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{2\sqrt{3}-\sqrt{2.3}}{2\sqrt{2}-2}=\dfrac{\sqrt{2.3}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{6}}{2}\)

c/ ĐKXĐ : \(a\ge0\)

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}=-\sqrt{a}\)

d/ \(p\ge0\)

\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)

\(\sqrt{x^2-8x+16}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\x^2-8x+16=\left(4-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\x^2-8x+16=16-8x+x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\0x=0\left(luônđúng\right)\end{matrix}\right.\)

Vậy \(x\le4\) là nghiệm của pt