a) Theo đề ta có :

\(A=\frac{3a-2b}{a-3b}\) với \(\frac{a}{b}=\frac{10}{3}\)

* \(\frac{a}{b}=\frac{10}{3}\) \(\Rightarrow a=\frac{10}{3}.b\)

Thay a = \(\frac{10b}{3}\) vào \(\frac{3a-2b}{a-3b}\)

\(\Rightarrow\frac{3a-2b}{a-3b}=\frac{3.\frac{10b}{3}-2b}{\frac{10b}{3}-3b}\) \(=\frac{10b-2b}{\frac{10b}{3}-\frac{9b}{3}}=\frac{8b}{\frac{b}{3}}=8b:\frac{b}{3}=8b.\frac{3}{b}=8.3=24\)

b) Theo đề ta có :

a - b = 3 => a = b + 3 

Thay a = b+3 vào \(B=\frac{a-8}{a-5}-\frac{4a-b}{3a+3}\)

\(\Rightarrow B=\frac{b+3-8}{b+3-5}-\frac{4.\left(b+3\right)-b}{3.\left(b+3\right)+3}\) \(=\frac{b-5}{b-2}-\frac{4b+12-b}{3b+9+3}=\frac{b-2-3}{b-2}-\frac{3b+12}{3b+12}\)

\(=\frac{b-2}{b-2}-\frac{3}{b-2}-1\) \(=1-\frac{3}{b-2}-1=0-\frac{3}{b-2}=-\frac{3}{b-2}\)

k đi!!!

a) \(\dfrac{4}{5}< \dfrac{5}{a}< \dfrac{10}{7}\) \(\left(a\inℤ\right)\)

\(\Leftrightarrow\dfrac{7}{10}< \dfrac{a}{5}< \dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{7.5}{10}< a< \dfrac{5}{4}.5\)

\(\Leftrightarrow\dfrac{7}{2}< a< \dfrac{25}{4}\)

\(\Leftrightarrow a\in\left\{4;5;6\right\}\)

b) \(\dfrac{2}{5}< \dfrac{a-1}{10}< \dfrac{8}{15}\)

\(\Leftrightarrow\dfrac{2.10}{5}< a-1< \dfrac{8.10}{15}\)

\(\Leftrightarrow4< a-1< \dfrac{16}{3}\)

\(\Leftrightarrow5< a< \dfrac{19}{3}\)

\(\Leftrightarrow a\in\left\{6\right\}\)

c) \(\dfrac{12}{7}< \dfrac{4}{a}< \dfrac{8}{3}\)

\(\Leftrightarrow\dfrac{3}{8}< \dfrac{a}{4}< \dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{3.4}{8}< a< \dfrac{7.4}{12}\)

\(\Leftrightarrow\dfrac{3}{2}< a< \dfrac{7}{3}\)

\(\Leftrightarrow a\in\left\{2\right\}\)

d) \(5< a^2-15< 16\)

\(\Leftrightarrow10< a^2< 31\)

\(\Leftrightarrow\sqrt[]{10}< a< \sqrt[]{31}\)

\(\Leftrightarrow a\in\left\{4;5\right\}\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

Bài 2: 

a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)

b: \(5A=5+5^2+...+5^{51}\)

\(\Leftrightarrow4A=5^{51}-1\)

hay \(A=\dfrac{5^{51}-1}{4}\)

Bài 3:

\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)

Ko ai biết làm sao?