a) \(\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}.\sqrt{2}}{2}=\frac{\sqrt{6}}{2}\)

b) \(\sqrt{\frac{3a}{5b}}=\frac{\sqrt{3a}}{\sqrt{5b}}=\frac{\sqrt{3a}.\sqrt{5b}}{5b}=\frac{\sqrt{15ab}}{5b}\left(a;b>0\right)\)

c) \(\sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{\sqrt{12}}=\frac{\sqrt{5}.\sqrt{12}}{12}=\frac{\sqrt{60}}{12}=\frac{2\sqrt{15}}{12}=\frac{\sqrt{15}}{6}\)

d) \(\sqrt{\frac{5x}{18y}}=\frac{\sqrt{5x}}{\sqrt{18y}}=\frac{\sqrt{5x}}{\sqrt{3^2.2y}}=\frac{\sqrt{5x}}{3\sqrt{2y}}\)

\(=\frac{\sqrt{5x}.\sqrt{3y}}{3.2y}=\frac{\sqrt{15xy}}{6xy}\)

Quên mất k ghi đk xy > 0

a) \(\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{2}\)                           b)\(\sqrt{\dfrac{3a}{5b}}=\dfrac{\sqrt{3a}.\sqrt{5b}}{\sqrt{5b}.\sqrt{5b}}=\dfrac{\sqrt{15ab}}{5b}\)       \(\sqrt{\dfrac{5}{12}}=\dfrac{\sqrt{5}.\sqrt{12}}{\sqrt{12}.\sqrt{12}}=\dfrac{\sqrt{60}}{12}\)                        d)             

Lời giải:

\(\sqrt{\frac{(1+\sqrt{2})^3}{27}}=\sqrt{\frac{(1+\sqrt{2})^3}{3^3}}=\sqrt{\frac{3(1+\sqrt{2})^3}{3^4}}\)

\(=\frac{(1+\sqrt{2})\sqrt{3+3\sqrt{2}}}{9}\)

\(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{(ab)^2(\frac{1}{a}+\frac{1}{b})}=\sqrt{ab^2+a^2b}\)

a: \(=6\cdot\sqrt{\dfrac{2xy}{4y^2}}\)

\(=6\cdot\dfrac{\sqrt{2xy}}{-2y}=-\dfrac{3\sqrt{2xy}}{y}\)

b: \(=\dfrac{4xy^2}{3}\cdot\dfrac{3}{\sqrt{xy}}=4\sqrt{x}\cdot y\sqrt{y}\)

mình mới học lớp 6 nên không biết

mày xem sẽ chưa tao xem rồi đó về mà xem

mất dạy thì biến nha

\(1,ĐKXĐ:x\ge0\\ x\sqrt{3}=-\sqrt{3x^2}\\ \Leftrightarrow3x^2=9x^2\\ \Leftrightarrow6x^2=0\\ \Leftrightarrow x=0\left(tm\right)\)

\(2,ab^2\sqrt{a}=ab^2\sqrt{a}\)

\(3,a\sqrt{\dfrac{b}{a}}=\sqrt{ab}\)

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