\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)

\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)

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

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

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

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

\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

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

\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

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

\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)

a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-3}{2}\)

\(=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{6^2}}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)

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

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

 \(VT\Leftrightarrow\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2=VP\left(dpcm\right)\)

\(\left(\dfrac{\sqrt{30}-\sqrt{20}}{\sqrt{3}-\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)

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

\(=5-3=2\)

a) -17√3/3                                                  b) 11√6 

c) 21                                                            d) 11

a)  326+236−426 và làm tiếp.
136x+6x):6x và làm tiếp

a) \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\)

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

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

\(=\frac{1}{6}\cdot\sqrt{6}=\frac{\sqrt{6}}{6}\left(đpcm\right)\)

b) \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)

\(=\left(\sqrt{6x}+\frac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)

\(=\left[\left(1+\frac{1}{3}+1\right)\sqrt{6x}\right]:\sqrt{6x}\)

\(=\frac{7}{3}\sqrt{6x}:\sqrt{6x}=\frac{7}{3}=2\frac{1}{3}\left(đpcm\right)\)

\(B=VT=\left(x\sqrt{\dfrac{6x}{x^2}}+\sqrt{\dfrac{6x}{3^2}}+\sqrt{6x}\right):\sqrt{6x}=\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}=\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}=\dfrac{7}{3}=2\dfrac{1}{3}=VP\Rightarrow\left(đpcm\right)\)

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

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

b: \(VT=\dfrac{\left(\sqrt{6x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right)}{\sqrt{6x}}\)

\(=1+\dfrac{1}{3}+1=2\dfrac{1}{3}\)

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Ta có đẳng thức : (2−3+√3√3+1).(2+3−√3√3−1)=1

xét vế trái ta có :(2−3+√3√3+1).(2+3−√3√3−1)  = 

a) ta co \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)=1\)

b) ta co \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)

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

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

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

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