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a: ĐKXĐ: a∉{-1/3;-3}

\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)

=>\(\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)

=>\(2\left(3a+1\right)\left(a+3\right)=\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)\)

=>\(2\left(3a^2+9a+a+3\right)=3a^2+9a-a-3+3a^2-9a+a-3\)

=>\(6a^2+20a+6=6a^2-6\)

=>20a=-12

=>a=-3/5(nhận)

b: ĐKXĐ: a∉{5/2;2/3}

\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)

=>\(\frac{2a-5-4}{2a-5}+\frac{3a-2+2}{3a-2}=2\)

=>\(1-\frac{4}{2a-5}+1+\frac{2}{3a-2}=2\)

=>\(\frac{2}{3a-2}=\frac{4}{2a-5}\)

=>\(\frac{4}{6a-4}=\frac{4}{2a-5}\)

=>6a-4=2a-5

=>4a=-1

=>a=-1/4(nhận)

c: ĐKXĐ: a<>-3

\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)

=>\(\frac{3a-1}{4a+12}+\frac{7a+2}{6a+18}=\frac{10}{3}-2=\frac43\)

=>\(\frac{3\left(3a-1\right)}{12\left(a+3\right)}+\frac{2\left(7a+2\right)}{12\left(a+3\right)}=\frac43\)

=>\(\frac{9a-3+14a+4}{12\left(a+3\right)}=\frac{4\cdot4\cdot\left(a+3\right)}{12\left(a+3\right)}\)

=>23a+1=16(a+3)=16a+48

=>7a=47

=>a=47/7(nhận)

2b: \(=8\sqrt{2}-3\sqrt{2}-3\sqrt{2}-10\sqrt{2}=-8\sqrt{2}\)

3:

a: \(=\left(\sqrt{6a}+\dfrac{\sqrt{6a}}{3}+\sqrt{6a}\right):\sqrt{6a}\)

=1+1/3+1

=7/3

b: \(=\dfrac{2}{3a-1}\cdot\sqrt{3}\cdot a\cdot\left|3a-1\right|\)

\(=\dfrac{2\sqrt{3}\cdot a\left(1-3a\right)}{3a-1}=-2a\sqrt{3}\)

\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)

\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)

\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)