Question

[Mai Chí Dũng]

a) \(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\)

b)Tính A= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{94\cdot97}+\dfrac{3}{97\cdot100}\)

Giúp mình với mình đang cần gấp .Cảm ơn

Answer 1

[Dang Tung]

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+1+\dfrac{x+6}{2004}+1+\dfrac{x+7}{2003}+1=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

b) \(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}+\dfrac{3}{97.100}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)