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

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

bạn làm rồi nên mk chỉ viết kq thôi nhé :)

a)\(\dfrac{4\sqrt{6x}}{3}\)

b)\(\left(2-y\right)\sqrt{xy}\)

Bài 2: 

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

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

\(\dfrac{\sqrt{x^6y^2}}{xy}=\dfrac{x^3y}{xy}=x^2\)

Lời giải:

\(A=\frac{(x-1)+(\sqrt{y}+\sqrt{xy})}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-\sqrt{y}}\\ =\frac{(\sqrt{x}-1)(\sqrt{x}+1)+\sqrt{y}(\sqrt{x}+1)}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-\sqrt{y}}\\ =\frac{(\sqrt{x}+1)(\sqrt{x}-1+\sqrt{y})}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-\sqrt{y}}\\ =\frac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}-\sqrt{y}}\)

\(A=\dfrac{x+\sqrt{y}+\sqrt{xy}-1}{\sqrt{x}+1}:\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\dfrac{\left(x-1\right)+\sqrt{y}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}:\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{y}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}:\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\dfrac{\left(\sqrt{x}-1+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)