(1/25.26 + 1/26.27+ ... + 1/ 29.30) . 150 + 103/100 : ( 103/10. 1/ x-1)
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1
15 tháng 7 2017
\(120\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120\left(\dfrac{1}{24}-\dfrac{1}{30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120.\dfrac{1}{120}+x=\dfrac{1}{3}\)
\(\Rightarrow1+x=\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{2}{3}\)
LL
2
23 tháng 6 2016
(1/24.25 + 1/25.26 + ... + 1/29.30) . 120 + x : 1/3 = -4
(1/24 - 1/25 + 1/25 - 1/26 + ... + 1/29 - 1/30) . 120 + x . 3 = -4
(1/24 - 1/30) . 120 + x . 3 = -4
(5/120 - 4/120) + x . 3 = -4
=> 1/120 . 120 + x . 3 = -4
=> 1 + x . 3 = -4
=> x . 3 = -4 - 1
=> x . 3 = -5
=> x = -5/3
Vậy x = -5/3
\(\left(\dfrac{1}{25.26}+\dfrac{1}{26.27}+......+\dfrac{1}{29.30}\right).150+\dfrac{103}{100}:\left(\dfrac{103}{10}.\dfrac{1}{x-1}\right)\)
\(=\left(\dfrac{1}{25}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{27}+....+\dfrac{1}{29}-\dfrac{1}{30}\right).150+\dfrac{103}{100}:\left(\dfrac{103}{10}.\dfrac{1}{x-1}\right)\)
\(=\left(\dfrac{1}{25}-\dfrac{1}{30}\right)\)\(.150+\dfrac{103}{100}:\left(\dfrac{103}{10}.\dfrac{1}{x-1}\right)\)
\(=\dfrac{1}{150}\)\(.150+\dfrac{103}{100}:\left(\dfrac{103}{10}.\dfrac{1}{x-1}\right)\)
\(=\dfrac{150}{150}+\dfrac{103}{100}:\left(\dfrac{103}{10}.\dfrac{1}{x-1}\right)\)
\(=1+\dfrac{103}{100}:\left(\dfrac{103}{10}.\dfrac{1}{x-1}\right)\)
`=203/100 : (103/10 . 1/(x-1) )`
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