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)

`Answer:`

a. Ta có: \(\frac{a}{b}=\frac{1}{3}\Rightarrow\frac{a}{1}=\frac{b}{3}\)

Đặt \(k=\frac{a}{1}=\frac{b}{3}\Rightarrow\hept{\begin{cases}a=k\\b=3k\end{cases}}\)

\(E=\frac{3a+2b}{4a-3b}\)

\(=\frac{3k+2.3k}{4k-3.3k}\)

\(=\frac{3k+6k}{4k-9k}\)

\(=\frac{9k}{-5k}\)

\(=-\frac{9}{5}\)

b. Thay `a-b=5` vào biểu thức `F`, ta được:

\(F=\frac{3a-\left(a-b\right)}{2a+b}-\frac{4b+\left(a-b\right)}{a+3b}\)

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

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

\(=1+1\)

\(=0\)

Câu 5:

\(D\left(2\right)=21a+9b-6a-4b\)

\(D\left(2\right)=\left(21a-6a\right)+\left(9b-4b\right)\)

\(D\left(2\right)=15a+5b\)

Mà: \(3a+b=18\Rightarrow b=18-3b\)

\(\Rightarrow D\left(2\right)=15a+5\left(18-3b\right)\)

\(D\left(2\right)=15a+90-15a\)

\(D\left(2\right)=90\)

Vậy: ...

còn câu 3, với 4 ạ?

4:

D=6a+9b=3(2a+3b)=36

5: 

D=15a+5b=5(3a+b)=90