5) ĐK: \(x\ge3\)

\(pt\Leftrightarrow2x+1=4+x-3+4\sqrt{x-3}\)

\(\Leftrightarrow x=4\sqrt{x-3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=16\left(x-3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-16x+48=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-4\right)\left(x-12\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=4\\x=12\end{matrix}\right.\end{matrix}\right.\left(tm\right)\)

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

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

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

\(M=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

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

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

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(Q=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)

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

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

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

\(=\dfrac{x}{\sqrt{x}-1}\)

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

\(\Leftrightarrow x-2\sqrt{x}+2>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

12) \(\sqrt{11+2\sqrt{30}}=\sqrt{\left(\sqrt{6}\right)^2+2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}\)

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

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

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

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

6) \(\dfrac{\sqrt{xy}-3\sqrt{y}}{\sqrt{y}\left(x-5\sqrt{x}+6\right)}=\dfrac{\sqrt{y}\left(\sqrt[]{x}-3\right)}{\sqrt{y}\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}\)

7) \(\left(2+\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(2+\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

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

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

\(4\sqrt{\dfrac{2}{9}}+\sqrt{2}+\sqrt{\dfrac{1}{18}}\)

\(=4\sqrt{2.\dfrac{1}{9}}+\sqrt{2}+\sqrt{\dfrac{2}{36}}\)

\(=\dfrac{4}{3}\sqrt{2}+\sqrt{2}+\dfrac{1}{6}\sqrt{2}\)

\(=\sqrt{2}\left(\dfrac{4}{3}+1+\dfrac{1}{6}\right)\)

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

Ta có: \(M=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)

\(\Rightarrow M=1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)

Dấu bằng xảy ra khi và chỉ khi: \(\sqrt{x}=0\Leftrightarrow x=0\)

KL:...

4) ĐK: \(x\ge3\)

pt <=> \(x^2-3x-2=x-3\)

\(\Leftrightarrow x^2-4x+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\left(n\right)\\x=2-\sqrt{3}\left(l\right)\end{matrix}\right.\)

1) ĐK: \(x\ge\dfrac{5}{2}\)

pt <=> \(x-4=\sqrt{2x-5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-4\right)^2=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-8x+16=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-10x+21=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-3\right)\left(x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left[{}\begin{matrix}x=3\left(l\right)\\x=7\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=7

2) ĐK: \(2x^2-8x+4\ge0\)

pt <=> \(\left\{{}\begin{matrix}x\ge2\\2x^2-8x+4=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\left(x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=0\left(l\right)\\x=4\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=4

3) ĐK: \(x\ge3\)

pt <=> \(\left\{{}\begin{matrix}x\le8\\x^2+x-12=x^2-16x+64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\17x=76\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=\dfrac{76}{17}\left(n\right)\end{matrix}\right.\) 

Vậy, pt có nghiệm duy nhất là \(x=\dfrac{76}{17}\)\(\)

b) \(\sqrt{\left(1-\sqrt{5}\right)^2}+\sqrt{23-8\sqrt{5}}\)

\(=\left|1-\sqrt{5}\right|+\sqrt{23-2\sqrt{60}}\)

\(=\sqrt{5}-1+\sqrt{\left(\sqrt{20}\right)^2-2.\sqrt{20}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{5}-1+\sqrt{\left(\sqrt{20}-\sqrt{3}\right)^2}\)

\(=\sqrt{5}-1+\sqrt{20}-\sqrt{3}=\sqrt{5}+2\sqrt{5}-1-\sqrt{3}\)

\(=3\sqrt{5}-1-\sqrt{3}\)