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\(C=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(C=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(C=\frac{2}{4\times5}+\frac{2}{5\times6}+\frac{2}{6\times7}+...+\frac{2}{15\times16}\)
\(C=2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(C=2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(C=2\times\frac{3}{16}=\frac{3}{8}\)
\(C=\frac{2}{20}+\frac{2}{30}+.........+\frac{2}{240}\)
\(=2\left(\frac{1}{4.5}+\frac{1}{5.6}+..........+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+.......+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2.\frac{3}{16}\)
\(=\frac{3}{8}\)
\(S=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
\(S=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\)
\(2S=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+....+\frac{1}{240}\)
\(2S=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{15.16}\)
\(2S=\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+.....+\left(\frac{1}{15}-\frac{1}{16}\right)\)
\(2S=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{15}-\frac{1}{16}\)
\(2S=\frac{1}{4}-\frac{1}{16}\)
\(2S=\frac{3}{16}\)
\(S=\frac{3}{8}\)
Ta có:
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2.\left(\frac{4}{16}-\frac{1}{16}\right)\)
\(=2.\frac{3}{16}=\frac{3}{8}\)
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
Ta có :
\(\frac{1}{10}< 1\)
\(\frac{1}{15}< 1\)
\(\frac{1}{21}< 1\)
........................
\(\frac{1}{120}< 1\)
\(\Rightarrow\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}< 1\)
\(\Rightarrow A< 1\)( đpcm)
Ta có : A = \(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
= \(\frac{1}{20}\times2+\frac{1}{30}\times2+...+\frac{1}{240}\times2\)
= \(2\times\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
= \(2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)
= \(2\times\frac{3}{16}\)
= \(\frac{3}{8}\)< 1
=> A < 1
Bạn nhân cả thừa số và mẫu số các phân số của tổng với 2 thì tổng ko thay đổi và ta được
2/20 + 2/30 + 2/42 + 2/56 + ... + 2/240
= 2/4 x 5 + 2/5 x 6 + 2/6 x 7 + 2/7 x 8 + ... + 2/15 x 16
= 2 x ( 1/4 - 1/5 + 1/5 - 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/15 - 1/16)
= 2 x (1/4 - 1/16)
= 2 x 3/16 = 3/8
Hoàng chúc bạn học tốt! Nếu mình đúng thì ấn thích nha!
\(C=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}=2\times\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(C=2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{15\times16}\right)\)
\(C=2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=2\times\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3}{8}\)
c = 2/20 + 2/30 +2/42 + ... 2 /240
=2/4.5 +2/5.6 + 2/6.7 + ... +1/15.16
=2. (1/4.5 + 1/5.6 + 1/6.7 +...+1/15.16)
= 2.(1/4-1/5+1/5-...-1/16)
=2.(1/4-1/6)=2.3/16=3/8.
ghi nho dau cham la dau nhan