\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

Để  \(P\) nguyên thì \(\frac{3}{\sqrt{x}+1}\) nguyên

\(\Rightarrow\sqrt{x}+1\in\left\{1;3;-1;-3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;-2;-4\right\}\)

\(\Rightarrow x\in\left\{0;\sqrt{2}\right\}\) 

\(\frac{4}{\sqrt{3-1}}+\frac{5}{3-\sqrt{2}}-2\sqrt{3}\)

\(=2\sqrt{3}+\frac{5}{3-\sqrt{2}}-2\sqrt{3}\)

\(=2\sqrt{2}+\frac{5\left(3+\sqrt{2}\right)}{7}-2\sqrt{3}\)

\(4\frac{4}{\sqrt{3-1}}+\frac{5}{3-\sqrt{2}}-2\sqrt{3}\)

\(=4\frac{4}{\sqrt{2}}+\frac{5}{3-\sqrt{2}}-2\sqrt{3}\)

\(=4+2\sqrt{2}+\frac{5}{3-\sqrt{2}}-2\sqrt{3}\)

\(=\frac{\left(4+2\sqrt{2}\right)\left(3-\sqrt{2}\right)+5-2\sqrt{3}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=\frac{12-4\sqrt{2}+6\sqrt{2}-4+5-6\sqrt{3}+2\sqrt{6}}{3-\sqrt{2}}\)

\(=\frac{13+2\sqrt{2}-6\sqrt{3}+2\sqrt{6}}{3-\sqrt{2}}\)

\(=\frac{\left(13+2\sqrt{2}-6\sqrt{3}+2\sqrt{6}\right)\left(3+\sqrt{2}\right)}{9-2}\)

\(=\frac{39+13\sqrt{2}+6\sqrt{2}+4-18\sqrt{3}-6\sqrt{6}+6\sqrt{6}+4\sqrt{3}}{7}\)

\(=\frac{43+19\sqrt{2}-14\sqrt{3}}{7}\)

\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)

Ta có

:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

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

\(=|2-\sqrt{5}|-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)

\(=-2=VP\left(đpcm\right)\)

\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)

Ta có:

\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)

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

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

\(=\frac{3+2\sqrt{2}}{2-1}\)

\(=3+2\sqrt{2}=VP\left(đpcm\right)\)

c,Bạn xem lại đề

\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)

Ta có:

\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)

\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)

\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)

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

\(=\frac{8}{5-4}\)

\(=8=VP\left(đpcm\right)\)

\(\sqrt{x^2-5}\ge0\Rightarrow x^2-5\ge0\)

\(\Rightarrow x^2\ge5\)

\(\Rightarrow x\ge\sqrt{5}\)

Vy Thị Hoàng Lan\(=-\sqrt{5}\)vẫn đúng nhé.

Ta có: \(\sqrt{x^2-5}=\sqrt{\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)}\)

Để căn thức có nghĩa thì \(x+\sqrt{5}\)và \(x-\sqrt{5}\)cùng dấu

\(TH1:\hept{\begin{cases}x+\sqrt{5}\ge0\\x-\sqrt{5}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-\sqrt{5}\\x\ge\sqrt{5}\end{cases}}\Leftrightarrow x\ge\sqrt{5}\)

\(TH1:\hept{\begin{cases}x+\sqrt{5}\le0\\x-\sqrt{5}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-\sqrt{5}\\x\le\sqrt{5}\end{cases}}\Leftrightarrow x\le-\sqrt{5}\)

a) \(\sqrt{1-4x+4x^2}=5\)

<=> \(\sqrt{4x^2-4x+1}=5\)

<=> 4x² - 4x + 1 = 5²

<=> 4x² - 4x + 1 = 25

<=> 4x² - 4x + 1 - 25 = 0

<=> 4x² - 4x - 24 = 0

<=> 4(x + 2)(x - 3) = 0

<=> x = -2 hoặc x = 3

 => x = -2 hoặc x = 3

b) \(\sqrt{4-5x}=12\)

<=> \(\sqrt{-5x+4}=12\)

<=> -5x + 4 = 12²

<=> -5x + 4 = 144

<=> -5x = 144 - 4

<=> -5x = 140

<=> x = -28

=> x = -28

\(a,\sqrt{1-4x+4x^2}=5\)

\(\Rightarrow4x^2-4x+1=25\)

\(\Rightarrow4x^2-4x-24=0\)

\(\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)

\(b,\sqrt{4-5x}=12\)

\(\Rightarrow4-5x=144\)

\(\Rightarrow5x=-140\)

\(\Rightarrow x=-28\)

\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)'

\(=|1+\sqrt{5}|+|1-\sqrt{5}|\)

\(=1+\sqrt{5}+\sqrt{5}-1\)

\(=2\sqrt{5}\)

\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

= \(\sqrt{5}+1+\sqrt{5}-1\)

= \(2\sqrt{5}\)

Giải hpt \(\hept{\begin{cases}\sin\alpha+\cos\alpha=\frac{7}{5}\\\sin^2\alpha+\cos^2\alpha=1\end{cases}}\) ra \(\hept{\begin{cases}\sin\alpha=\frac{4}{5}\\\cos=\frac{3}{5}\end{cases}}\)\(\Rightarrow\)\(\tan\alpha=\frac{4}{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(4.\frac{\sqrt{6}}{2+\sqrt{6}}=\frac{\sqrt{6}.4}{2+\sqrt{6}}=4\left(3-\sqrt{6}\right)\)

\(A=4x^2+4x-6|2x+1|+6.\)

\(=\left(4x^2+4x+1\right)-6|2x+1|+5\)

\(=\left[\left(2x+1\right)^2-2.|2x+1|.3+9\right]-4\)

\(=\left(|2x+1|-3\right)^2-4\)

Vì \(\left(|2x+1|-3\right)^2\ge0\Rightarrow\left(|2x+1|-3\right)^2-4\ge-4\)Hay \(A\ge-4\)

Vậy giá trị nhỏ nhất của A=-4 , Dấu '=' xảy ra khi \(|2x+1|-3=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)