Tìm x , biết \(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=1\frac{1929}{1991}\)
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KN
5 tháng 5 2019
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(\frac{8}{3}+2-\frac{2}{x+1}=1\frac{1989}{1991}\)
\(\frac{2}{x+1}=\frac{13}{10}\)( số thập phân dài quá nên mk lấy số tròn thôi nha )
\(x+1=2:\frac{13}{10}\)
\(x+1=\frac{20}{13}\)
\(\Leftrightarrow x=\frac{7}{13}\)
18 tháng 8 2020
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(1+\frac{1}{3}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{x+1-x}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x-1}=\frac{1990}{1991}\)
\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x-1}=\frac{1990}{1991}\)
\(\frac{1}{x-1}=\frac{11}{6}-\frac{1990}{1991}=\frac{9961}{11946}\)
\(x-1=\frac{11946}{9961}\Rightarrow x=\frac{21907}{9961}\)
DP
13 tháng 7 2017
\(1+\frac{1}{3}+\frac{1}{6}+....+\frac{2}{x\left(x+1\right)}=4\)
\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{x\left(x+1\right)}=4\)
\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{x\left(x+1\right)}=4\)
\(\Leftrightarrow1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)\right]=4\)
\(\Leftrightarrow1+2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=4\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{4-1}{2}=\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{3}{2}=-1\)
\(\Leftrightarrow x=-1+1=-2\)
Vậy x = -2
DP
16 tháng 7 2017
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1992\)
18 tháng 3 2019
Câu 1:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}.\)
\(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)
\(\frac{1}{2.3}.2+\frac{1}{3.4}.2+\frac{1}{4.5}.2+...+\frac{1}{x.\left(x+1\right)}.2=\frac{1991}{1993}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
...
e tự tính nốt nha
KN
19 tháng 3 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{1993}\div2\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1993-1\)
\(\Leftrightarrow x=1992\)
Vậy x = 1992
25 tháng 2 2019
a)Ta có \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)