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=> 1/2+1/6+1/12+1/20+....+1/x.(x+1) = 1992/1993
=> 1/2+1/2.3+1/3.4+1/4.5+.....+1/x.(x+1) = 1992/1993
=> 1/2+1/2-1/3+1/3-1/4+1/4-1/5+.....+1/x-1/x+1 = 1992/1993
=> 1 - 1/x+1 = 1992/1993
=> x/x+1 = 1992/1993
=> x = 1992
Vậy x = 1992
Tk mk nha
\(\Rightarrow\frac{2}{2}+\frac{2}{2.3}+\frac{2}{2.6}+...+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)
\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{3984}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3984}{1993}:2\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{x}{x+1}=\frac{1992}{1993}\)
=>x=1992
chuyển 1 đi còn cái kia=1991/1993
nhân mỗi p/số với 1/2 rồi p/tích mẫu=2.3,3.4........x.(x+1)
lập hiệu ra rồi tính OK???
=2/2+2/6+2/12+...+...+2/x(x+1)=3984/1993
=2(1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=3984/1993
=2(1-1/2+1/2-1/3+14-1/5+...+1/x-1/x+1=3984/1993
=2(1-1/x+1)=3984/1993
=2-2/x+1=3984/1993
2/x+1=2/1993
x+1=1003
x=1992
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)
=> \(\frac{1}{1.\left(1+1\right):2}+\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+\frac{1}{4.\left(4+1\right):2}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)
=> \(\frac{1}{\frac{1.\left(1+1\right)}{2}}+\frac{1}{\frac{2.\left(2+1\right)}{2}}+...+\frac{1}{\frac{x.\left(x+1\right)}{2}}=1\frac{1991}{1993}\)
=> \(\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{x.\left(x+1\right)}=1\frac{1991}{1993}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{1991}{1993}\)
=> \(1-\frac{1}{x+1}=1\frac{1991}{1993}\)
=> \(\frac{1}{x+1}=\frac{-1991}{1993}\)
=> -1991.(x + 1) = 1993
=> -1991x - 1991 = 1993
=> -1991x = 3984
=> x = \(-\frac{3984}{1991}\)
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
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{1993}\div2\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1993-1\)
\(\Leftrightarrow x=1992\)
Vậy x = 1992
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 \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
