Question

Cho D=\(\dfrac{5}{6X37}\)+\(\dfrac{1}{6X43}\)+\(\dfrac{6}{7X43}\)+\(\dfrac{10}{7X59}\)

và E=\(\dfrac{8}{9X37}\)+\(\dfrac{2}{9X47}\)+\(\dfrac{3}{10X47}\)+\(\dfrac{9}{10X59}\)

Em hãy tính \(\dfrac{D}{E}\)

giúp minh với mai mình phải nộp rùi

Answer 1

=>\(D=7\left(\dfrac{5}{42\cdot37}+\dfrac{1}{42\cdot43}+\dfrac{6}{43\cdot49}+\dfrac{10}{49\cdot59}\right)\)

\(=7\left(\dfrac{1}{37}-\dfrac{1}{42}+\dfrac{1}{42}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{59}\right)\)

=7(1/37-1/59)

=7*22/2183

\(E=5\left(\dfrac{8}{37\cdot45}+\dfrac{2}{45\cdot47}+\dfrac{3}{47\cdot50}+\dfrac{9}{50\cdot59}\right)\)

\(=5\left(\dfrac{1}{37}-\dfrac{1}{45}+\dfrac{1}{45}-\dfrac{1}{47}+...+\dfrac{1}{50}-\dfrac{1}{59}\right)\)

=5(1/37-1/59)

=>D/E=7/5