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\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
\(\Rightarrow x+1=2017\)
\(\Rightarrow x=2017-1=2016\)
Vậy x = 2016
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{2016}{2017}\)
1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)
\(\dfrac{3}{4}\)+\(\dfrac{1}{x\left(x+1\right)}\)=\(\dfrac{2016}{2017}\)
\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{2013}{8068}\)
Bn tự lm tiếp nhé!!! Sorry mk đang vội
11: \(=\left(1+\dfrac{1}{98}-1-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)=0\)
12: \(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\left(\dfrac{-6+5}{10}\right)^2\)
\(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\dfrac{1}{100}=\dfrac{7}{17}+\dfrac{1}{170}=\dfrac{71}{170}\)
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
\(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)..............\left(1-\dfrac{2}{99.100}\right)\)
\(=\left(\dfrac{6}{2.3}-\dfrac{2}{2.3}\right).\left(\dfrac{12}{3.4}-\dfrac{2}{3.4}\right)..............\left(\dfrac{9900}{99.100}-\dfrac{2}{99.100}\right)\)
\(=\dfrac{4}{2.3}.\dfrac{10}{3.4}..........................\dfrac{9898}{99.100}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}..............\dfrac{98.101}{99.100}\)
\(=\dfrac{1.2.3.....98}{2.3......99}.\dfrac{4.5.6.....101}{3.4.......100}\)
\(=\dfrac{1}{99}.\dfrac{101}{3}=\dfrac{101}{297}\)
\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)
\(\left(x+1\right)^2=324=18^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=18\\x+1=-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-19\end{matrix}\right.\)
Ta có \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{x+1}{324}+\dfrac{1}{x+1}\)
\(\Rightarrow\)\(\dfrac{1}{x+1}-\dfrac{x+1}{324}=0\)
\(\Rightarrow\)\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)
\(\Rightarrow\)(x+1).(x+1)=324
\(\Rightarrow\)(x+1)2=324
\(\Rightarrow\)(x+1)2 = 182 = (-18)2
TH1: (x+1)2 = 182
\(\Rightarrow\)x+1 = 18
\(\Rightarrow\)x = 17
TH2: (x+1)2 = (-18)2
\(\Rightarrow\)x+1 = -18
\(\Rightarrow\)x = -19
Vậy x\(\in\)\(\left\{17;-19\right\}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{600}\)
=> x + 1 = 600
x = 600 - 1
x = 599
Vậy x = 599