\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\)\(\cdot\dfrac{12}{3^2}\cdot\dfrac{20}{4^2}\cdot\cdot\cdot\cdot\dfrac{110}{10^2}\)\(\cdot\left(x-2\right)=-20\left(x+1\right)+60\)
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27 tháng 6 2022
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
PN
13 tháng 7 2017
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
25 tháng 5 2022
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)
\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)
=>83/12x=5/12
hay x=5/83
29 tháng 8 2022
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
20 tháng 6 2018
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
2 tháng 1 2021
Ta có: \(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\left(\dfrac{6}{5}+\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\dfrac{8}{5}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{32}{65}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\left(2x+1\right)^2=\dfrac{10}{13}:\dfrac{32}{65}=\dfrac{10}{13}\cdot\dfrac{65}{32}=\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{5}{4}\\2x+1=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{4}-1=\dfrac{1}{4}\\2x=-\dfrac{5}{4}-1=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}:2=\dfrac{1}{8}\\x=-\dfrac{9}{4}:2=-\dfrac{9}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{1}{8};-\dfrac{9}{8}\right\}\)
\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2