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Ta có:
\(A=\frac{112}{13.20}+\frac{112}{20.27}+.........+\frac{112}{62.69}\)
\(\Rightarrow A=112.\left(\frac{1}{13.20}+\frac{1}{20.27}+..........+\frac{1}{62.69}\right)\)
\(\Rightarrow A=16.\left(\frac{7}{13.20}+\frac{7}{20.27}+.......+\frac{7}{62.69}\right)\)
\(\Rightarrow A=16.\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+........+\frac{1}{62}-\frac{1}{69}\right)\)
\(\Rightarrow A=16.\left(\frac{1}{13}-\frac{1}{69}\right)\)
\(\Rightarrow A=16.\frac{56}{897}\)
\(\Rightarrow A=\frac{896}{897}\)
Vậy: \(A=\frac{896}{897}\)
\(A=\frac{112}{13.20}+\frac{112}{20.27}+\frac{112}{27.34}+...+\frac{112}{62.69}\)
\(\Rightarrow A=112.\left(\frac{1}{13.20}+\frac{1}{20.27}+\frac{1}{27.34}+...+\frac{1}{62.69}\right)\)
\(\Rightarrow A=112.\frac{7}{7}.\left(\frac{1}{13.20}+\frac{1}{20.27}+\frac{1}{27.34}+...+\frac{1}{62.69}\right)\)
\(\Rightarrow A=112.\frac{1}{7}\left(\frac{7}{13.20}+\frac{7}{20.27}+\frac{7}{27.34}+...+\frac{7}{62.69}\right)\)
\(\Rightarrow A=16\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+\frac{1}{27}-\frac{1}{34}+...+\frac{1}{62}-\frac{1}{69}\right)\)
\(\Rightarrow A=16\left(\frac{1}{13}-\frac{1}{69}\right)\)
\(\Rightarrow A=16.\frac{56}{897}\)
\(\Rightarrow A=\frac{896}{897}\)
\(Can\)\(you\) \(k\) \(for\) \(me,everyone?\)
a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100