Question

Rút gọn các biểu thức sau:

a. \(ab^2.\sqrt{\dfrac{3}{a^2b^4}}\) với a < 0, \(b\ne0;\)                            b. \(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}\) với a > 3;

c. \(\sqrt{\dfrac{9+12a+4a^2}{b^2}}\) với \(a\ge-1,5\) và b < 0;      

d. \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\) với a < b <0.

Answer 1

a,\(ab^2\sqrt{\dfrac{3}{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{ab^2}=\sqrt{3}\)

b,\(\sqrt{\dfrac{27\left(a-3\right)^2}{48}}=\dfrac{3\sqrt{3}\left(a-3\right)}{4\sqrt{3}}=\dfrac{3}{4}\left(a-3\right)\)

c,\(\sqrt{\dfrac{9+12a+4a^2}{b^2}}=\dfrac{\sqrt{\left(3+2a\right)^2}}{\sqrt{b^2}}=\dfrac{3+2a}{b}\)

d, \(\left(a-b\right).\sqrt{\dfrac{ab}{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\sqrt{\left(a-b\right)^2}}=\left(a-b\right).\dfrac{\sqrt{ab}}{\left(a-b\right)}=\sqrt{ab}\)