Question

cho 4a-b=6. Tính \(\dfrac{6a-b}{3a+5}-\dfrac{4a-4b}{3b-5}\)

Answer 1

4a-b=6 nên b=4a-6

\(\dfrac{6a-b}{3a+5}-\dfrac{4a-4b}{3b-5}\)

\(=\dfrac{6a-\left(4a-6\right)}{3a+5}-\dfrac{4a-4\left(4a-6\right)}{3\left(4a-6\right)-5}\)

\(=\dfrac{6a-4a+6}{3a+5}-\dfrac{4a-16a+24}{12a-18-5}\)

\(=\dfrac{2a+6}{3a+5}-\dfrac{-12a+24}{12a-23}\)

\(=\dfrac{2a+6}{3a+5}+\dfrac{12a-24}{12a-23}\)

\(=\dfrac{\left(2a+6\right)\left(12a-23\right)+\left(12a-24\right)\left(3a+5\right)}{\left(3a+5\right)\left(12a-23\right)}\)

\(=\dfrac{24a^2-46a+72a-138+36a^2+60a-72a-120}{\left(3a+5\right)\left(12a-23\right)}\)

\(=\dfrac{60a^2+14a-258}{\left(3a+5\right)\left(12a-23\right)}\)