\(\dfrac{2x}{4}+\dfrac{2x}{28}+\dfrac{2x}{70}+...+\dfrac{2x}{418}=\dfrac{14}{11}\)
\(\dfrac{2x}{1.4}+\dfrac{2x}{4.7}+\dfrac{2x}{7.10}+\dfrac{2x}{10.13}+...+\dfrac{2x}{19.22}=\dfrac{14}{11}\)
Ta có công thức : \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức ta có :
\(\dfrac{2x}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}\right)=\dfrac{14}{11}\)
\(\dfrac{2x}{3}.\left(1-\dfrac{1}{22}\right)=\dfrac{14}{11}\)
\(\Leftrightarrow\) \(\dfrac{2x}{3}.\left(\dfrac{21}{22}\right)=\dfrac{14}{11}\)
\(\Rightarrow\) \(\dfrac{2x}{3}=\dfrac{14}{11}:\dfrac{21}{22}=\dfrac{4}{3}\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Ai thấy đúng thì ủng hộ nha !!!!
\(\dfrac{X}{2}+\dfrac{X}{14}+\dfrac{X}{35}+\dfrac{X}{65}+\dfrac{X}{104}+\dfrac{X}{209}=1\dfrac{3}{11}\\ X\cdot\left(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{209}\right)=\dfrac{14}{11}\\ X\cdot\left(\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{418}\right)=\dfrac{14}{11}\\ X\cdot\dfrac{2}{3}\cdot\left(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{418}\right)=\dfrac{14}{11}\\ \dfrac{2}{3}X\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}\right)=\dfrac{14}{11}\\ \dfrac{2}{3}X\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\right)=\dfrac{14}{11}\\ \dfrac{2}{3}X\cdot\left(1-\dfrac{1}{19}\right)=\dfrac{14}{11}\\ \dfrac{2}{3}X\cdot\dfrac{18}{19}=\dfrac{14}{11}\\ \dfrac{2}{3}X=\dfrac{14}{11}:\dfrac{18}{19}\\ \dfrac{2}{3}X=\dfrac{14}{11}\cdot\dfrac{19}{18}\\ \dfrac{2}{3}X=\dfrac{133}{99}\\ X=\dfrac{133}{99}:\dfrac{2}{3}\\ X=\dfrac{133}{99}\cdot\dfrac{3}{2}\\ X=\dfrac{133}{66}\)