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

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

b/ \(\sqrt{x-1}+\sqrt{x+1}=\sqrt{7}\) (ĐKXĐ: ...)

\(\Leftrightarrow2x+2\sqrt{x^2-1}=7\)

\(\Leftrightarrow2\sqrt{x^2-1}=7-2x\) (\(x\le\frac{7}{2}\))

\(\Leftrightarrow4\left(x^2-1\right)=\left(7-2x\right)^2\)

\(\Leftrightarrow28x=53\)

\(\Leftrightarrow x=\frac{53}{28}\)

đk : x ≥ 2 
Bạn bình phương 2 vế, thu gọn đc: 
3√[x(x−2)(x+1)] ≤ 2x2−6x−2 
<=> 3√[(x2−2x)(x+1)] ≤ 2(x2−2x) − 2(x+1) 
Chia 2 vế cho (x+1), đặt t= căn((x2−2x)/(x+1)), t≥ 0 ta đc: 
2t^2 - 3t - 2 ≥ 0 => t ≥ 2 
<=> x^2 - 2x ≥ 4x + 4 
<=> x^2 - 6x -4 ≥ 0 
<=> x ≥ 3+√13

P/s: Tham khảo nhé

\(\sqrt{x+2\sqrt{x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

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

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

\(=\sqrt{x+2.\left|\sqrt{x}-2\right|}+\sqrt{x-2.\left|\sqrt{2x}-2\right|}\)

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

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

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

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

\(=2x+2\sqrt{x}-2\sqrt{2x}\)

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

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

\(A=\frac{5-\sqrt{5}}{\sqrt{5}-1}=\frac{5\sqrt{5}+5-5-\sqrt{5}}{\sqrt{5^2}-1}=\frac{5\sqrt{5}-\sqrt{5}}{5-1}=\frac{4\sqrt{5}}{4}=\sqrt{5}\)

\(M=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{2x+2\sqrt{x}+3\sqrt{x}+3}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}.\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2018}{\sqrt{x}+1}\)

\(\frac{\sqrt{x}+2018}{\sqrt{x}+1}=1+\frac{2017}{\sqrt{x}+1}\le2018\)

Dấu "=" xảy ra \(\Leftrightarrow\)

... 

Ta có:\(x\ge\sqrt{2}\Rightarrow x^2\ge2\Rightarrow\sqrt{x^2-1}-1\ge0\) (*)

\(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)

\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)

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

Kết hợp với (*), ta có:

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

Vậy ...

a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)

b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)

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

\(=\left(6-5\sqrt{2}\right)x\)

c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)