Bài 1: Rút gọn :
a) \(\sqrt{16ab^2}\) : \(\sqrt{4a}\) với a > 0 , b < 0
b) \(\sqrt{52a^3}\) : \(\sqrt{13a}\) với a > 0
c) \(\dfrac{2\sqrt{a^2\left(1-a\right)^2}}{\sqrt{50}}\) với a > 1
d) \(\dfrac{\sqrt{3ab^4}}{\sqrt{27a}}\) với a > 0
e) \(\dfrac{\sqrt{125a}}{\sqrt{5a}}\) với a > 0
f) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}\) với x > 0
a:
Vì b<0 nên |b|=-b
\(\sqrt{16ab^2}:\sqrt{4a}=\sqrt{\dfrac{16ab^2}{4a}}=\sqrt{4b^2}=2\left|b\right|=-2b\)
b: a>0 nên |a|=a
\(\sqrt{52a^3}:\sqrt{13a}=\sqrt{\dfrac{52a^3}{13a}}=\sqrt{4a^2}=\sqrt{4}\cdot\sqrt{a^2}\)
\(=2\cdot\left|a\right|=2a\)
c:
a>1 nên a-1>0
=>|a-1|=a-1
\(2\cdot\dfrac{\sqrt{a^2\left(1-a\right)^2}}{\sqrt{50}}\)
\(=2\cdot\left|a\right|\cdot\dfrac{\left|1-a\right|}{5\sqrt{2}}\)
\(=\dfrac{2}{5\sqrt{2}}\cdot a\left(a-1\right)=\dfrac{\sqrt{2}}{5}\cdot\left(a^2-a\right)\)
d: \(\dfrac{\sqrt{3ab^4}}{\sqrt{27a}}=\sqrt{\dfrac{3ab^4}{27a}}=\sqrt{\dfrac{b^4}{9}}=\dfrac{\sqrt{b^4}}{\sqrt{9}}=\dfrac{b^2}{3}\)
e: a>0
=>|a|=a
\(\dfrac{\sqrt{125a}}{\sqrt{5a}}=\sqrt{\dfrac{125a}{5a}}=\sqrt{25}=5\)
f:
Vì x>0 nên |x|=x
\(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\sqrt{\left(\dfrac{4}{x}\right)^2}=\left|\dfrac{4}{x}\right|=\dfrac{4}{x}\)
