Cho biểu thức $B = \dfrac{{\sqrt x  + 3}}{{\sqrt x  + 2}}$với $x \ge 0$. So sánh $B$ với $1$.

$\sqrt {\dfrac{A}{{BC}}}  = \dfrac{{\sqrt {ABC} }}{B}$

$\sqrt {\dfrac{A}{B}}  =  - \dfrac{{\sqrt {ABC} }}{{BC}}$           

$\sqrt {\dfrac{A}{{BC}}}  = \dfrac{{\sqrt {ABC} }}{{BC}}$

$\sqrt {\dfrac{A}{{BC}}}  = \dfrac{{ABC}}{{\sqrt {BC} }}$

$3 + 4\sqrt 2$

$4$

$2$

$4\sqrt 2$

$9\left( {2 - y} \right)$

$81{\left( {2 - y} \right)^2}$

$9{\left( {2 - y} \right)^2}$

$- 9{\left( {2 - y} \right)^2}$

$\sqrt {5{y^2}}$

$\sqrt {25{y^3}}$

$\sqrt {5{y^3}}$

$\sqrt {25y\sqrt y }$

$16\sqrt 2  + 12\sqrt 3$

$15\sqrt 3$

$12\sqrt 3$

$16\sqrt 2$

$\dfrac{{\sqrt 6 }}{6}$

$\sqrt 6$

$\dfrac{{\sqrt 6 }}{2}$

$\dfrac{{\sqrt 6 }}{3}$