Question

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

B=\(\dfrac{a-8}{b-5}-\dfrac{4a-b}{3a+3}\) với a-b=3,b\(\ne\)5,a\(\ne\)-1

Giúp mình với

Answer 1

a, Theo bài ta có :

\(\dfrac{a}{b}=\dfrac{10}{3}\Leftrightarrow\dfrac{a}{10}=\dfrac{b}{3}\)

Đặt :

\(\dfrac{a}{10}=\dfrac{b}{3}=k\left(k\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=10k\\b=3k\end{matrix}\right.\)

Ta có :

\(Q=\dfrac{3a-2b}{a-3b}=\dfrac{3.10k-2.3k}{10k-3.3k}=\dfrac{30k-6k}{10k-9k}=\dfrac{24k}{1k}=24\)

Vậy ...........

Answer 2

a-b=3=>a=b+3 Thay a=b+3 vào B

\(\Rightarrow B=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\)

\(\Rightarrow B=1-\dfrac{4b-b+12}{3b+9+3}=1-1=0\)

Answer 3

\(\dfrac{a}{b}=\dfrac{10}{3}\Rightarrow3a=10b\Rightarrow a=\dfrac{10}{3}b\)

thay vào A ta có:

\(A=\dfrac{10b-2b}{\dfrac{10}{3}b-3b}=\dfrac{8b}{\dfrac{1}{3}b}=8:\dfrac{1}{3}=24\)