Question

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

Answer 1

Ta có : \(\frac{4a-b}{3a+5}=\frac{3a+\left(a-b\right)}{3a+5}=\frac{3a+5}{3a+5}=1\)

            \(\frac{3b-a}{2b-5}=\frac{2b+b-a}{2b-5}=\frac{2b-a+b}{2b-5}=\frac{2b-\left(a-b\right)}{2b-5}=\frac{2b-5}{2b-5}=1\)

Nên : \(\frac{4a-b}{3a+5}+\frac{3b-a}{2b-5}=1+1=2\)

Answer 2

có nhiều cách, có thể là cách này

a-b=5 => a=b+5

=> \(\frac{4a-b}{3a+5}+\frac{3b-a}{2b-5}=\frac{4\left(b+5\right)-b}{3\left(b+5\right)+5}+\frac{3b-\left(b+5\right)}{2b-5}=\frac{4b+20-b}{3b+15+5}+\frac{3b-b-5}{2b-5}\)

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

Answer 3

thay 5=a-b vào