ĐK: \(x>0,x\ne4\)

a) \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b) Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}+1\)

\(\Rightarrow M=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)

c) Do \(\sqrt{x}>0\). Để \(M>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\) (TM)

\(13^8:13^4=13^{8-4}=13^4=28561\)

Với \(a\ge b\ge0\) \(\Rightarrow\sqrt{a}\ge\sqrt{b}\Rightarrow\sqrt{a}-\sqrt{b}\ge0\)

a) \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\Leftrightarrow a+b+2\sqrt{ab}\ge a+b\Leftrightarrow2\sqrt{ab}\ge0\) (luôn đúng)

b) \(\sqrt{a}-\sqrt{b}\le\sqrt{a-b}\Leftrightarrow a+b-2\sqrt{ab}\le a-b\Leftrightarrow2b\le2\sqrt{ab}\)

\(\Leftrightarrow b\le\sqrt{ab}\Leftrightarrow\sqrt{b}\le\sqrt{a}\) (luôn đúng)

Vậy ta có điều phải chứng minh

Điều kiện: \(x\ge1\)

\(\sqrt{x+2\sqrt{x-1}}=2\\ \Rightarrow\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=2\\ \Rightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\\ \Rightarrow\sqrt{x-1}+1=2\\ \Rightarrow\sqrt{x-1}=1\\ \Rightarrow x-1=1\\ \Rightarrow x=2\left(TM\right)\)

\(\sqrt{x^2}=7\Rightarrow\left|x\right|=7\Rightarrow x=\pm7\)