Đáp án:
a) $\sqrt{35} < 6 <\sqrt{42} < 7$
b) $\dfrac52$
Giải thích các bước giải:
a) Ta có:
$6 = \sqrt{36}$
$7 =\sqrt{49}$
Do $35 < 36 < 42 < 49$
nên $\sqrt{35} < \sqrt{36} < \sqrt{42} < \sqrt{49}$
hay $\sqrt{35} < 6 <\sqrt{42} < 7$
b) $\left(\dfrac{\sqrt{216}}{3} -\dfrac{\sqrt6 -2\sqrt3}{\sqrt8 - 2}\right)\cdot\dfrac16$
$= \left(\dfrac{\sqrt{2^3.3^3}}{3} -\dfrac{\sqrt6 -\sqrt2\sqrt6}{2\sqrt2 - 2}\right)\cdot\dfrac{1}{\sqrt6}$
$= \left(\dfrac{6\sqrt6}{3} -\dfrac{\sqrt6(1 - \sqrt2)}{2(\sqrt2 - 1)}\right)\dfrac{1}{\sqrt6}$
$= \left(2\sqrt6 +\dfrac{\sqrt6}{2}\right)\cdot\dfrac{1}{\sqrt6}$
$= \dfrac{2\sqrt6}{\sqrt6} +\dfrac{\sqrt6}{2\sqrt6}$
$= 2 +\dfrac12=\dfrac52$
