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\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=18-1\)
\(\Leftrightarrow x=17\)
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)
\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)
\(\Leftrightarrow x=\pm\frac{29}{12}\)
A=\(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)=\(\frac{1}{7}-\frac{1}{14}\)=\(\frac{1}{14}\)
B=0
\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)
\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)
\(\text{Đề }\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\left(1-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)
=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)
=> 27x - 27 = 30x - 10
=> 27x - 30x = -10 + 27
=> -3x = 17
=> x = -17/3.
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\frac{1}{x+1}=\frac{1}{2017}\)
\(\Rightarrow\)x+1=2017
\(\Rightarrow\)x=2017-1
x=2016
Vậy x=2016
Chúc bạn học tốt+-*/
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{40341}=\frac{1}{2017}\)
\(\Rightarrow x+1=2017\Rightarrow x=2016\)
2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2.[1/6+1/12+1/20+...+1/x.(x+1)]=2013/2015
1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2013/4030
1/2-1/3+1/3-1/4+...+1/x-1/x+1=2013/4030
1/2-1/x+1=2013/4030
1/x+1=1/2015
=> x+1=2015
x=2014
Vậy x=2014
Đặt A=Vế trái
Ta có :
\(A \over 2 \)\(= \)\({1\over 6 } +{1\over 12 }+{1\over 20 }+...+{1\over x(x+1)}\)
=\({1\over 2}-{1\over 3}+{1\over 3}-{1\over 4}+{1\over4}-{1\over 5}+...+{1\over x-1}-{1\over x}+{1\over x}-{1\over x+1}\)
=\({1\over2}-{1\over x+1}\)
Từ đó suy ra: \({1\over2}-{1\over x+1}={2013\over4030}\)
=> x=2014
caau hoi tuong tu cungx cos