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!!!
