Cho $Q = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}$. Tìm $x$ để $Q = 3$

$\sqrt {\dfrac{a}{b}}  = \dfrac{{\sqrt a }}{{\sqrt b }}$

$\dfrac{{\sqrt {ab} }}{{\sqrt c }} = \sqrt {\dfrac{{ab}}{c}}$

$\dfrac{{\sqrt a }}{{\sqrt {bc} }} = \dfrac{{\sqrt {ab} }}{{\sqrt c }}$

Cả A, B đều đúng.

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

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

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

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

$32$

$16$

$64$

$8$

$\dfrac{{1,1}}{{240}}$

$\dfrac{{11}}{{24}}$

$\dfrac{{11}}{{240}}$

$\dfrac{{240}}{{11}}$

$\dfrac{{25}}{{27}}$    

$- \dfrac{{25}}{{27}}$

$- \dfrac{5}{7}$

Không tồn tại.

$- 33$

$- 132$

$132$

Không tồn tại.

$3a{\left( {4a - 3} \right)^3}$

$- 3a{\left( {4a - 3} \right)^3}$

$3a\left( {4a - 3} \right)$

$3a{\left( {3 - 4a} \right)^3}$