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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
a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\frac{99}{100}-x=\frac{2}{3}\)
\(x=\frac{99}{100}-\frac{2}{3}\)
\(x=\frac{97}{300}\)
b) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)
\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)
\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)
\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)
\(\Rightarrow\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)
\(\Rightarrow\frac{99}{100}-x=\frac{2}{3}\)
\(\Rightarrow x=\frac{99}{100}-\frac{2}{3}\)
\(\Rightarrow x=\frac{97}{300}\)
b, k hiểu đề :v
a) = 99x + (1+2+3+4+...+99)=0
99x+4950=0
99x=0-4950
99x=-4950
x=-4950:99
x=-50
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.
\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)-x+\frac{221}{231}=\frac{4}{3}\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)
\(\left(\frac{21-11}{231}\right)+\frac{221}{231}-x=\frac{4}{3}\)
\(\frac{10}{231}+\frac{221}{231}-x=\frac{4}{3}\)
\(\frac{231}{231}-x=\frac{4}{3}\)
\(1-x=\frac{4}{3}\)
\(x=1-\frac{4}{3}\)
\(x=\frac{3-4}{3}\)
\(x=-\frac{1}{3}\)
Chúc bạn học tốt
Tìm x :
a) (x - 3) + (x - 2) + (x - 1) + .... + 10 + 11 = 11
(x - 3) + (x - 2) + (x - 1) + .... + 10 = 0
[(x - 3) + (x - 2) + (x - 1)] + (0 + 1 + 2 + ... + 10) = 0
[(x - 3) + (x - 2) + (x - 1)] + 55 = 0
x - 3 + x - 2 + x - 1 = -55
x + x + x - (3 + 2 + 1) = -55
x3 - 6 = -55
x3 = -55 + 6
x3 = -49
x = -49 : 3
x = -\(\frac{49}{3}\)
\(A=\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{55}{334}=\dfrac{11}{334}\)
\(B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{20}{21}\)