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)             

a: \(=-xy\cdot\dfrac{\sqrt{xy}}{x}=-y\sqrt{yx}\)

b: \(=\sqrt{\dfrac{-105x^3}{35^2}}=\sqrt{-105x}\cdot\dfrac{x}{35}\)

c: \(=\sqrt{\dfrac{5a^3b}{49b^2}}=\sqrt{5ab}\cdot\dfrac{a}{7b}\)

d: \(=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3}\cdot\sqrt{xy}\)

mình mới học lớp 6 nên không biết

mày xem sẽ chưa tao xem rồi đó về mà xem

mất dạy thì biến nha

) x y \sqrt{\dfrac{x}{y}}=x y \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x y}{y} \sqrt{x y}=x \sqrt{x y} .xyyx​​=xyy2xy​​=yxy​xy​=xxy​.

b) \dfrac{x}{y} \sqrt{\dfrac{x}{y}}=\dfrac{x}{y} \sqrt{\dfrac{x y}{y^{2}}}=\dfrac{x}{y^{2}} \sqrt{x y} .yx​yx​​=yx​y2xy​​=y2x​xy​.

c) \sqrt{\dfrac{1}{a}+\dfrac{1}{a^{2}}}=\sqrt{\dfrac{a+1}{a^{2}}}=\dfrac{\sqrt{a+1}}{a} .a1​+a21​​=a2a+1​​=aa+1​​.

d) \sqrt{\dfrac{4 x^{3}}{25 y}}=\sqrt{\dfrac{4 x^{2} x y}{25 y^{2}}}=\dfrac{2 x}{5 y} \sqrt{x y} .

a, GTTĐ x nhân căn xy

b,x/y^2 nhân xăn xy

c,  căn( a+1)/a

d, 2x căn (xy)/5y

e, 2 nhân căn (3ab) 

a) x căn xy

b) x/y^2 căn xy

c) căn a+1/a

d) 2x/5y căn xy

e) 2 căn 3ab

a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)

\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)

\(=-7\sqrt{3}\cdot\sqrt{xy}\)

b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

$a)-7xy.\sqrt{\dfrac{3}{xy}}$

$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$

$=-7.\sqrt{\dfrac{xy}{3}}$

$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$

$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$

$=(\sqrt{a}+1)(b\sqrt{a}+1)$

a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)

b) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)

b: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

Bài 2: 

a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)

b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)