Tìm x biết: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)\times x}=\frac{15}{16}\)
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\(\text{Đề }\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\left(1-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)
=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)
=> 27x - 27 = 30x - 10
=> 27x - 30x = -10 + 27
=> -3x = 17
=> x = -17/3.
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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2020}\)
\(\Rightarrow x+1=2020\Leftrightarrow x=2019\)
Vậy x = 2019
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\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\frac{3}{2}=1\)
\(\Leftrightarrow3x=-\frac{1}{2}\)
\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)
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
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ta có 1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=17/18
1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=17/18
1-1/x+1=17/18
1/x+1=1-17/18
1/x+1=1/18
suy ra: x+1=18
x=18-1
x=17
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\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
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a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)
\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)
\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)
\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)
=> x = 4
b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)
\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)
\(\Rightarrow100x-99x=1-\frac{1}{100}\)
\(\Rightarrow x=\frac{99}{100}\)
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\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right).\frac{1}{x}=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(\frac{8+4+2+1}{16}.\frac{1}{x}=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\left(1-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\frac{15}{16}\)
=> \(\frac{1}{x}=3\)
=> \(x=\frac{1}{3}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)\times x}=\frac{15}{16}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}=\frac{15}{16}\)
\(1-\frac{1}{x}=\frac{15}{16}\)
\(\frac{1}{x}=\frac{1}{16}\)
\(\Rightarrow x=16\)