a, \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)

do \(a\ge0\)

b, \(\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{\frac{676a}{a}}=\sqrt{676}=26\)

c, \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\left|15a\right|-3a\)

\(=15a-3a=12a\)do a > 0 

d, \(=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

\(=\left(3-a\right)^2-\sqrt{36a^2}=\left(3-a\right)^2-\left|6a\right|\)

Với \(a\ge0\Rightarrow\left(3-a\right)^2-6a=a^2-6a+9-6a=a^2-12a+9\)

Với \(a< 0\Rightarrow\left(3-a\right)^2+6a=a^2-6a+9+6a=a^2+9\)

a) Ta có:

b) Ta có:

c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:

d) Ta có:

b) \(\sqrt{13a}\).\(\sqrt{\frac{52}{a}}\)=\(\sqrt{13a.\frac{52}{a}}\)=\(\sqrt{13.13.2.2}\)=13.2=26

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)

c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

a. \(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}=\sqrt{\dfrac{3a.2a}{2.75}}=\sqrt{\dfrac{3a^2}{75}}=\sqrt{\dfrac{a^2}{25}}=\dfrac{\sqrt{a^2}}{\sqrt{25}}=\dfrac{a}{5}\)

b.\(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}=\sqrt{5a}.\sqrt{2}=\sqrt{10a}\)

a.\(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}=\dfrac{\sqrt{3a}}{\sqrt{2}}.\dfrac{\sqrt{2a}}{\sqrt{25}.\sqrt{3}}=\dfrac{a}{5}\) b. \(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}=\dfrac{\sqrt{5}.\sqrt{a}.\sqrt{2a}}{\sqrt{a}}=\sqrt{10a}\)

a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)

\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)

\(=\sqrt{\dfrac{6a^2}{24}}\)

\(=\sqrt{\dfrac{a^2}{4}}\)

\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)

\(=\dfrac{a}{2}\)

b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)

\(=\sqrt{3a\cdot\dfrac{52}{a}}\) 

\(=\sqrt{3\cdot52}\)

\(=\sqrt{13\cdot3\cdot4}\)

\(=2\sqrt{39}\)

c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)

\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)

\(=2y^2\cdot\dfrac{x^2}{-2y}\)

\(=\dfrac{2y^2\cdot x^2}{-2y}\)

\(=-x^2y\)

đk : \(a\ge0;b\ge0;a\ne b\)

a) \(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

= \(\dfrac{a+2\sqrt{ab}+b+a-2\sqrt{ab}+b}{a-b}\) = \(\dfrac{2\left(a+b\right)}{a-b}\)

b) đk : \(a\ge0;b\ge0;a\ne b\)

\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

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

= \(\dfrac{\sqrt{a}+\sqrt{b}}{1}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}\)

= \(\dfrac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{a+b}\)

\(\sqrt{\dfrac{2a}{3}.}\sqrt{\dfrac{3a}{8}=\sqrt{\dfrac{2a}{3}.\sqrt{\dfrac{3a}{8}}}=\sqrt{\dfrac{2.a}{3.8}}}\)

\(=\sqrt{\dfrac{\left(2.3\right)\left(a.a\right)}{3.8}=\sqrt{\dfrac{6a^2}{24}}}\)

\(=\sqrt{\dfrac{6a^2}{6.4}}=\sqrt{\dfrac{a^2}{4}=}=\sqrt{\dfrac{a^2}{2^2}}\)

\(=\sqrt{\dfrac{a}{2}}^2=\dfrac{a}{2}\)

Vì \(a>0\) nên \(\dfrac{a}{2}>0\)\(=\dfrac{a}{2}\)

\(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}.Với,a\ge0,Ta,Có,\dfrac{\sqrt{2a}}{\sqrt{3}}\cdot\dfrac{\sqrt{3a}}{\sqrt{8}}=\dfrac{\sqrt{2a}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\cdot\dfrac{\sqrt{3a}\cdot\sqrt{8}}{\sqrt{8}\cdot\sqrt{8}}=\dfrac{\sqrt{6a}}{3}\cdot\dfrac{\sqrt{24a}}{8}=\dfrac{\sqrt{6a}\cdot\sqrt{24a}}{3\cdot8}=\dfrac{\sqrt{144a^{^2}}}{24}=\dfrac{\sqrt{\left(12a\right)^{^2}}}{24}=\dfrac{\left|12a\right|}{24}=\dfrac{12a}{24}=\dfrac{a}{2}\)

a: \(B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{x-9}=\dfrac{x+9}{x-9}\)

b: \P=A:B

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{x-9}{x+9}=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x+9}>=\dfrac{-1\cdot3}{9}=\dfrac{-1}{3}\)

Dấu = xảy ra khi x=0