1/5.9 +1/8.11 +1/11.14 +….+ 1/x+(x+5) =101/1540 x+1/3 + 1/6 + 1/10+….+ 2/x.(x+1)=1 Làm ơn ik mà mik đang cần gấp

0 bình luận về “1/5.9 +1/8.11 +1/11.14 +….+ 1/x+(x+5) =101/1540 x+1/3 + 1/6 + 1/10+….+ 2/x.(x+1)=1 Làm ơn ik mà mik đang cần gấp”

1. $a/\dfrac{1}{5.8}+\dfrac{1}{8.11}+….+\dfrac{1}{x(x+5)}=\dfrac{101}{1540}$

   $⇔\dfrac{1}{5}.5.(\dfrac{1}{5.8}+\dfrac{1}{8.11}+….+\dfrac{1}{x(x+5)})=\dfrac{101}{1540}$

   $⇔\dfrac{1}{5}.(\dfrac{5}{5.8}+\dfrac{5}{8.11}+….+\dfrac{5}{x(x+5)})=\dfrac{101}{1540}$

   $⇔\dfrac{1}{5}.(\dfrac{1}{5}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-……+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{101}{1540}$

   $⇔\dfrac{1}{5}.(\dfrac{1}{5}-\dfrac{1}{x+5})=\dfrac{101}{1540}$

   $⇒\dfrac{1}{5}-\dfrac{1}{x+5}=\dfrac{505}{1540}$

   $⇔\dfrac{1}{x+5}=\dfrac{1}{5}-\dfrac{505}{1540}=\dfrac{-197}{1540}$

   $⇒x+5=\dfrac{-1540}{197}$

   $⇒x=\dfrac{-2525}{197} (???)$

   b/

   $x+\dfrac{1}{3}+ \dfrac{1}{6} + \dfrac{1}{10}+….+ \dfrac{2}{x.(x+1)}=1$

   $⇔\dfrac{1}{3}+ \dfrac{1}{6} + \dfrac{1}{10}+….+ \dfrac{2}{x.(x+1)}= 1 – x$

   $⇔\dfrac{2}{6} + \dfrac{2}{12} + \dfrac{2}{20} + … + \dfrac{2}{x(x + 1)} = 1 – x$

   $⇔2(\dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + … + \dfrac{1}{x(x + 1)} = 1 – x$

   $⇔\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} +\dfrac{ 1}{4} – \dfrac{1}{5} + … + \dfrac{1}{x} – \dfrac{1}{x + 1} = \dfrac{1 – x}{2}$

   $⇔\dfrac{1}{2} – \dfrac{1}{x + 1)} = \dfrac{(1 – x}{2}$

   $⇔\dfrac{(x + 1 – 2}{2.(x + 1} = \dfrac{(1 – x}{2}$

   $⇒ (\dfrac{x – 1}{x + 1} = 1-x$

   $⇔ \dfrac{1}{x + 1} = -1$

   $⇔ x + 1 = -1 $

   $⇒ x = -2$

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