Ta có: \(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}}{3}+\frac{\sqrt{2}}{3}+\frac{2\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12}.\sqrt{\frac{1}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}.\sqrt{12\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)}\)

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

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

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

\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left(\sqrt{3}-\sqrt{2}\right)\)(vì \(\sqrt{3}-\sqrt{2}>0\))

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

\(\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\frac{3+6\sqrt{3}}{\sqrt{3}}-\frac{13}{\sqrt{3}+4}\)

\(=\frac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\frac{4^2-\left(\sqrt{3}\right)^2}{\sqrt{3}+4}\)

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

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

Ta có: \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)

\(=\sqrt{\frac{\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}{\left(5-\sqrt{21}\right)^2}}+\sqrt{\frac{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}{\left(5+\sqrt{21}\right)^2}}\)

\(=\sqrt{\frac{4}{\left(5-\sqrt{21}\right)^2}}+\sqrt{\frac{4}{\left(5+\sqrt{21}\right)^2}}\)

\(=2\left(\frac{1}{5-\sqrt{21}}+\frac{1}{5+\sqrt{21}}\right)\)

\(=2.\frac{5+\sqrt{21}+5-\sqrt{21}}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}=\frac{2.10}{4}=5\)

x - 3 = ( √x )² - ( √3 )² = ( √x - √3 )( √x + √3 ) < với x > 0 >

            Bài làm :

Ta có :

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

Giải thích các bước giải:

   Ta có:

     \(\Delta ADB\)và  \(\Delta\widehat{ADB}=90^o\)

    \(\Rightarrow sinA=\frac{BD}{AB}\left(1\right)\)

      \(\hept{\begin{cases}\widehat{BEH}=\widehat{BDA}=90^o\\\widehat{B}chung\end{cases}}\)

   \(\Rightarrow\Delta BEH~\Delta BDA\left(g.g\right)\)

   \(\Rightarrow\frac{BE}{BD}=\frac{BH}{BA}\)

   \(\Rightarrow\frac{BE}{BH}=\frac{BD}{BA}\) 

  Khi đó:

       \(\hept{\begin{cases}\widehat{B}chung\\\frac{BE}{BH}=\frac{BD}{BA}\end{cases}}\)

   \(\Rightarrow\Delta BED~\Delta BHA\left(g.g\right)\)

   \(\Rightarrow\frac{ED}{HA}=\frac{BD}{BA}\left(2\right)\)

  \(\left(1\right),\left(2\right)\Rightarrow\frac{ED}{HA}=sinA\)

giúp mình với mình đang cần gấp :((

a)\(\text{ĐKXĐ: }a\ne4;a>0\)

b)\(\text{Đặt BT là A, ta có: }\)

\(A=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)+4\sqrt{a}-4}{4-a}\)

\(A=\frac{\left(a+5\sqrt{a}+6\right)-\left(a-\sqrt{a}-2\right)+4\sqrt{a}-4}{4-a}\)

\(A=\frac{10\sqrt{a}+4}{4-a}\)

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

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

  3\(x\) - y

= (\(\sqrt{3x}\))² - (\(\sqrt{y}\))²

= (\(\sqrt{3x}\) - \(\sqrt{y}\)).(\(\sqrt{3x}\) + \(\sqrt{y}\))