Question

bài 25:rút gọn các biểu thức sau

a) \(\sqrt{3a}\).\(\sqrt{27a}\) với a\(\ge\)0

b)\(\sqrt{8a}\).\(\sqrt{\dfrac{32}{a}}\) với a>0

c)\(\sqrt{2x}\).\(\sqrt{3x}\).\(\sqrt{6x^2}\)

d)\(\sqrt{3x}\).\(\sqrt{\dfrac{48}{x}}\)

e)\(\dfrac{1}{a-1}\).\(\sqrt{9\left(a-1\right)^2}\) với a>1

Answer 1

a/ \(\sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=3\sqrt{9a^2}=3.3.a=9a\) ( vì \(a\ge0\) )

b/ \(\sqrt{8a}.\sqrt{\dfrac{32}{a}}=2\sqrt{2a}.\dfrac{4\sqrt{2a}}{\sqrt{a}}=2\sqrt{2}.4\sqrt{2a}=8\sqrt{4a}=16\sqrt{a}\)

c/ĐK: x\(\ge\)0

\(\sqrt{2x}.\sqrt{3x}.\sqrt{6x^2}=\sqrt{2x}.\sqrt{3x}.\left|x\right|\sqrt{2}.\sqrt{3}\) = \(x.2\sqrt{x}.3\sqrt{x}\) ( vì x\(\ge0\) ) = \(6x^2\)

d/ Đk: x>0

\(\sqrt{3x}.\sqrt{\dfrac{48}{x}}=\sqrt{3}.\sqrt{x}.\dfrac{4\sqrt{3}}{\sqrt{x}}=4.3=12\)

e/ \(\dfrac{1}{a-1}.\sqrt{9\left(a-1\right)^2}=\dfrac{1}{a-1}.\sqrt{9}.\sqrt{\left(a-1\right)^2}\)

= \(\dfrac{1}{a-1}.3.\left|a-1\right|=\dfrac{1}{a-1}.3.\left(a-1\right)\) ( vì a>1)

= 3