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Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Mặt khác:
\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)
\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)
\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)
\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)
Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Mặt khác:
\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)
\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)
\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)
\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)
\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)
\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)
\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)
\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)
\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)
\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)
\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)
I don't now
mik ko biết
sorry
......................
b,\(B=2^2+4^2+...+20^2\)
\(\Rightarrow B=2^2\left(1^2+2^2+...+10^2\right)\)
\(\Rightarrow B=4.\left[1.\left(2-1\right)+2.\left(3-1\right)+...+10.\left(11-1\right)\right]\)
\(\Rightarrow B=4\left(1.2-1+2.3-2+...+10.11-10\right)\)
\(\Rightarrow B=4\left[\left(1.2+2.3+...+10.11\right)-\left(1+2+...+10\right)\right]\)
\(\Rightarrow B=4\left(\frac{10.11.12}{3}-\frac{11.10}{2}\right)\)
a)Ta có:
A= 1/1.2+1/2.3+1/3.4+.....+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b)Ta có:
B= 1/11+1/12+1/13+1/14+1/15+...+1/50
=(1/11+1/50)+(1/12+1/49)+...+(1/30+1/31)
=61/11.50+61/12.49+...+61/30.31
=61.(1/11.50+1/12.49+...+1/30.31)
Mình xin lỗi chỉ làm được đến đây vì dạng tính B mình không tốt lắm ◕◡◕
\(B=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)>\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)=> \(B>\frac{20}{30}+\frac{20}{50}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>1\)
mà \(A=\frac{99}{100}
