Question

Cho a-b = 5. Tính \(\dfrac{4a-b}{3a+5}+\dfrac{3b-a}{2b-5}\)

Answer 1

Giải:
Ta có: \(a-b=5\Leftrightarrow a=b+5\)

\(\dfrac{4a-b}{3a+5}+\dfrac{3b-a}{2b-5}=\dfrac{4b+20-b}{3b+15+5}+\dfrac{3b-b-5}{2b-5}\)

\(=\dfrac{3b+20}{3b+20}+\dfrac{2b-5}{2b-5}=1+1=2\)

Vậy...

Answer 2

ta có : a-b=5 => a=b+5 khi đó pt trên trở thành:

\(\dfrac{3a+a-b}{3a+5}+\dfrac{2b+b-a}{2b+5}=\dfrac{3a+5}{3a+5}+\dfrac{2b+5}{2b+5}=1+1=2\)

vậy ......