Tìm x, biết : ( 11/12 + 11/11.12 + 11/23.24 + ........ + 11/99.100) + x = 2/3 . Ai đó lm ơn giúp với
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Mình nghĩ câu A phải là 11/23.34 mới đúng.
Theo đề mới: A)
11/12 = 11/1.12
Vậy A = 11/1.12 + 11/12.23 + ... + 11/89.100
= 1 - 1/12 + 1/12 - 1/23 + .... + 1/89 - 1/100
= 99/1000
Vậy x là: 2/3 - 99/100 = -97/300
B) 2/11.13 + ... + 2/19.21 = 1/11 - 1/13 + .... + 1/19 - 1/21 = 10/231
=> 10/231 - x = 4/3 - 221/223 = 229/669
=> x = 10/231 - 229/669
=> x = 6690/154539 - 52899/154539
=> x = -46209/154539 = -15403/51513
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a.
\(\left(\frac{11}{12}+\frac{11}{12\times23}+\frac{11}{23\times34}+...+\frac{11}{89\times100}\right)+x=\frac{2}{3}\)
\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{2}{3}\)
\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\)
\(\left(\frac{12}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\)
\(\left(1-\frac{1}{100}\right)+x=\frac{2}{3}\)
\(\left(\frac{100-1}{100}\right)+x=\frac{2}{3}\)
\(\frac{99}{100}+x=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{99}{100}\)
\(x=\frac{200-297}{300}\)
\(x=-\frac{97}{300}\)
b.
\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)-x+\frac{221}{231}=\frac{4}{3}\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)
\(\left(\frac{21-11}{231}\right)+\frac{221}{231}-x=\frac{4}{3}\)
\(\frac{10}{231}+\frac{221}{231}-x=\frac{4}{3}\)
\(\frac{231}{231}-x=\frac{4}{3}\)
\(1-x=\frac{4}{3}\)
\(x=1-\frac{4}{3}\)
\(x=\frac{3-4}{3}\)
\(x=-\frac{1}{3}\)
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