a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)

\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)

\(Q=\dfrac{1}{x-2\sqrt{x}+3}\)

Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)

a) \(x>0,x\ne1\)

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

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

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

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

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

c) \(P< 0\Rightarrow\dfrac{x-1}{\sqrt{x}}< 0\) mà \(\sqrt{x}>0\Rightarrow x-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)

a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)

b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)

\(=2\sqrt{2}+\sqrt{3}\)

c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)

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

d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)

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

\(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

\(\Rightarrow2A=2\sqrt[3]{2+\sqrt{5}}+2\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)

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

\(\Rightarrow A=1\)

\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

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

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

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

\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)

a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

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

\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)

a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)

\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)

b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)

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

c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)

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

d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)

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

Kẻ đường cao BH của tam giác 

\(\Rightarrow S_{ABC}=\dfrac{1}{2}.AC.BH=\dfrac{1}{2}.\dfrac{BH}{AB}.AC.AB=\dfrac{1}{2}.sinA.AC.AB\)

\(=\dfrac{1}{2}.b.c.sinA\)

\(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\left(x>0,x\ne1\right)\)

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

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

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

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

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

b) \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=1+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge1+2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=3\)

\(\Rightarrow P\ge3\)

Xét \(P=3\Rightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Rightarrow x=1\) (trái với ĐKXĐ)

\(\Rightarrow P>3\)

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

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

\(=\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}{2}}-\sqrt{6}\)

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

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

\(=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}-\sqrt{6}=\dfrac{2\sqrt{3}}{\sqrt{2}}-\sqrt{6}=\sqrt{6}-\sqrt{6}=0\)

\(F=\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)\)

\(=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}-\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}-2\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}-2=\left|\sqrt{7}+1\right|-\left|\sqrt{7}-1\right|-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\)

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

\(=2\sqrt{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{2.\dfrac{5}{2}}\right)\)

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

\(=2\sqrt{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)=2\sqrt{6}\)

\(H=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}\)

\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\)

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

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