Rút gọn biểu thức  $\dfrac{{{a^2}}}{{11}}.\sqrt {\dfrac{{121}}{{{a^4}{b^{10}}}}}$ với $ab \ne 0$ ta được:

$\dfrac{{MN}}{{NP}}$

$\dfrac{{MP}}{{NP}}$

$\dfrac{{MN}}{{MP}}$

$\dfrac{{MP}}{{MN}}$

$BH = 2\,cm$, $CH = 3,2\,cm$, $AC = 4\,cm$, $AH = 2,4\,cm$

$BH = 1,8\,cm$, $CH = 3,2\,cm$, $AC = 4\,cm$, $AH = 2,4\,cm$.

$BH = 1,8\,cm$, $CH = 3,2\,cm$, $AC = 3\,cm$, $AH = 2,4\,cm$

$BH = 1,8\,cm$, $CH = 3,2\,cm$, $AC = 4\,cm$, $AH = 4,2\,cm$

$x = 14$

$x = 13$

$x = 12$

$x = \sqrt {145}$

${b^2} = b'.a$

$\dfrac{1}{{{h^2}}} = \dfrac{1}{{{c^2}}} + \dfrac{1}{{{b^2}}}$ 

$a.h = b'.c'$

${h^2} = b'.c'$

$A{H^2} = BH.CH$

$A{B^2} = BH.BC$

$\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}$

$AH.AB = BC.AC$

$a > b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}$

$a < b \Leftrightarrow \sqrt[3]{a} < \sqrt[3]{b}$

$a \ge b \Leftrightarrow \sqrt[3]{a} \ge \sqrt[3]{b}$

$a < b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}$

$\sqrt {2018 + 2019}  = \sqrt {2018}  + \sqrt {2019}$

$\sqrt {2018. 2019}  = \dfrac{{\sqrt {2018} }}{{\sqrt {2019} }}$

$\sqrt {2018} .\sqrt {2019}  = \sqrt {2018.2019}$

$2018. 2019 = \dfrac{{\sqrt {2019} }}{{\sqrt {2018} }}$