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câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
\(\Leftrightarrow x+1=2017\Leftrightarrow x=2016\)
Vậy \(x=2016\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
\(\Rightarrow x+1=2017\)
\(\Rightarrow x=2017-1=2016\)
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
a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)
\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)
\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)
\(x=\dfrac{-9198}{4400}\)
a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(x+\dfrac{206}{100}=5\)
\(x=5-\dfrac{206}{100}\)
\(x=\dfrac{147}{50}\)
Vậy \(x=\dfrac{147}{50}\)
Lời giải:
a.
$32^{10}=(2^5)^{10}=2^{50}$
$8^{16}=(2^3)^{16}=2^{3.16}=2^{48}$
Vì $2^{50}> 2^{48}$ nên $32^{10}> 8^{16}$
b.
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{x(x+1)}=\frac{-19}{20}\)
\(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{x+1-x}{x(x+1)}=\frac{-19}{20}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{-19}{20}\)
\(1-\frac{1}{x+1}=\frac{-19}{20}\)
\(\frac{x}{x+1}=\frac{-19}{20}\)
Khả năng $\frac{-19}{20}$ là $\frac{19}{20}$. Khi đó $x=19$.