Khử mẫu của biểu thức lấy căn
\(a,\sqrt{\dfrac{-2}{3a^2}}\) (a<0)
\(b,\sqrt{\dfrac{1}{200}}\)
\(c,\sqrt{\dfrac{7}{500}}\)
\(d,\sqrt{\dfrac{3}{98}}\)
\(e,\sqrt{\dfrac{\left(1-\sqrt{2}\right)^2}{8}}\)
\(f,a\sqrt{\dfrac{1}{a}}\left(a>0\right)\)
\(g,\sqrt{\dfrac{4a^3}{64b}}\left(a,b< 0\right)\)
\(h,2ab\sqrt{\dfrac{3}{ab}}\left(ab>0\right)\)
a, Vì trong dấu căn là số âm nên biểu thức này vô nghĩa. b)\(\sqrt{\dfrac{1}{200}}=\dfrac{1}{\sqrt{200}}=\dfrac{1}{10\sqrt{2}}=\dfrac{\sqrt{2}}{10\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{2}}{20}\)
c,\(\sqrt{\dfrac{7}{500}}=\dfrac{\sqrt{7}}{\sqrt{500}}=\dfrac{\sqrt{7}}{10\sqrt{5}}=\dfrac{\sqrt{7}.\sqrt{5}}{10\sqrt{5}.\sqrt{5}}=\dfrac{\sqrt{35}}{50}\)
d, \(\sqrt{\dfrac{3}{98}}=\dfrac{\sqrt{3}}{\sqrt{98}}=\dfrac{\sqrt{3}}{\sqrt{49.2}}=\dfrac{\sqrt{3}}{7\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{7\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{14}\)
e, \(\sqrt{\dfrac{\left(1-\sqrt{2}\right)^2}{8}}=\dfrac{\left|1-\sqrt{2}\right|}{\sqrt{8}}=\dfrac{\sqrt{2}-1}{2\sqrt{2}}=\dfrac{\sqrt{2}.\left(\sqrt{2}-1\right)}{2\sqrt{2}.\sqrt{2}}=\dfrac{2-\sqrt{2}}{4}\)
f,\(a\sqrt{\dfrac{1}{a}}=a\dfrac{1}{\sqrt{a}}=\dfrac{\left(\sqrt{a}\right)^2}{\sqrt{a}}=\sqrt{a}\) g, \(a,b< 0\Rightarrow\sqrt{\dfrac{4a^3}{64b}}xácđịnh,có:\sqrt{\dfrac{4a^3}{64b}}=\dfrac{2\sqrt{a^3}}{8\sqrt{b}}=\dfrac{\sqrt{a^3b}}{4\sqrt{b}.\sqrt{b}}=\dfrac{a\sqrt{ab}}{4b}\)
h, ab>0=>\(\sqrt{\dfrac{3}{ab}}\) có nghĩa, ta có:\(2ab\sqrt{\dfrac{3}{ab}}=2ab\dfrac{\sqrt{3}}{\sqrt{ab}}=\dfrac{2\sqrt{3}ab}{\sqrt{ab}}=2\sqrt{3ab}\)
