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Question

tìm x:2/1.6+2/6.11+2/11.16+...+2/x=41/103

HT.Phong (9A5) · Accepted Answer

$\dfrac{2}{1\cdot6}+\dfrac{2}{11\cdot16}+...+\dfrac{2}{x\left(x+5\right)}=\dfrac{41}{103}\ =>\dfrac{2}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{x\left(x+5\right)}\right)=\dfrac{41}{103}\
=>1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{41}{103}:\dfrac{2}{5}=\dfrac{205}{206}\
=>1-\dfrac{1}{x+5}=\dfrac{205}{206}\
=>1-\dfrac{1}{x+5}=1-\dfrac{1}{206}\
=>\dfrac{1}{x+5}=\dfrac{1}{206}\
=>x+5=206\
=>x=206-5=201$Câu trả lời: $\dfrac{2}{1\cdot6}+\dfrac{2}{11\cdot16}+...+\dfrac{2}{x\left(x+5\right)}=\dfrac{41}{103}\ =>\dfrac{2}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{x\left(x+5\right)}\right)=\dfrac{41}{103}\
=>1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{41}{103}:\dfrac{2}{5}=\dfrac{205}{206}\
=>1-\dfrac{1}{x+5}=\dfrac{205}{206}\
=>1-\dfrac{1}{x+5}=1-\dfrac{1}{206}\
=>\dfrac{1}{x+5}=\dfrac{1}{206}\
=>x+5=206\
=>x=206-5=201$

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