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Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)
=> \(S< \frac{3}{4}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\frac{3}{2}=1\)
\(\Leftrightarrow3x=-\frac{1}{2}\)
\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)
Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Leftrightarrow x=99\)
a) => ( x + 1/2 ) . 3 = 1
=> 3x + 3/2 = 1
=> 3x = 1 - 3/2
=> 3x = -1/2
=> x = -1/2 : 3 = -1/6
\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow a=\frac{2020}{2019}\)
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\)
=> \(A=\frac{9}{10}\)
b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> \(A=1+\frac{7}{n-5}\)
Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)
=> n=(-2; 4, 6, 8)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{6}{7}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{6}{7}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{6}{7}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{6}{7}=\frac{1}{7}\)
\(\Rightarrow x+1=7\)
\(\Rightarrow x=7-1=6\)
vậy x = 6
\(\frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{x\left(x+1\right)}=\frac{6}{7}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...........-\frac{1}{x}-\frac{1}{x+1}=\frac{6}{7}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{6}{7}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{7}\)
\(\Rightarrow x+1=7\)
\(\Rightarrow x=6\)
Vậy x = 6
Ta có:
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{20}\)
\(=1-\frac{1}{x+1}=\frac{19}{20}\)
\(\frac{1}{x+1}=1-\frac{19}{20}\)
\(\frac{1}{x+1}=\frac{1}{20}\)
\(x+1=20\)
\(x=19\)
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{x\left(x+1\right)}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{x+1-x}{x\left(x+1\right)}=\frac{19}{20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{19}{20}\)
\(=1-\frac{1}{\left(x+1\right)}=\frac{19}{20}\)
\(=\frac{1}{\left(x+1\right)}=1-\frac{19}{20}=\frac{1}{20}\)
\(\Rightarrow x=\frac{\left(20-1\right)}{1}=19\)
Vậy \(x=19\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}\right).x=\frac{23}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}:\frac{1}{2}\)
\(\left(\frac{1}{2}-\frac{1}{9.10}\right).x=\frac{46}{45}\)
\(\frac{22}{45}.x=\frac{46}{45}\)
\(x=\frac{46}{45}:\frac{22}{45}\)
\(x=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
Ta có:\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{9.10}\right)=\frac{1}{2}\cdot\frac{22}{45}=\frac{11}{45}\)
=>\(\frac{11}{45}\cdot x=\frac{23}{45}\)
=>11x=23
=>x=23/11
\(1+\frac{1}{3}+\frac{1}{6}+....+\frac{2}{x\left(x+1\right)}=4\)
\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{x\left(x+1\right)}=4\)
\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{x\left(x+1\right)}=4\)
\(\Leftrightarrow1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)\right]=4\)
\(\Leftrightarrow1+2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=4\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{4-1}{2}=\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{3}{2}=-1\)
\(\Leftrightarrow x=-1+1=-2\)
Vậy x = -2
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1992\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
\(1-\frac{1}{x+1}=\frac{44}{45}\)
\(\frac{1}{x+1}=1-\frac{44}{45}\)
\(\frac{1}{x+1}=\frac{1}{45}\)
=> x + 1 = 45
x = 45 - 1
x = 44