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1 / 12 + 1 / 20 + 1 / 30 + 1 / 42 + 1 / 56 + 1 / 72 + 1 / 90 + 1 / 110 + 1 / 132 =
(2 310 * 1) / (2 310 * 12) + (1 386 * 1) / (1 386 * 20) + (924 * 1) / (924 * 30) + (660 * 1) / (660 * 42) + (495 * 1) / (495 * 56) + (385 * 1) / (385 * 72) + (308 * 1) / (308 * 90) + (252 * 1) / (252 * 110) + (210 * 1) / (210 * 132) =
2 310 / 27 720 + 1 386 / 27 720 + 924 / 27 720 + 660 / 27 720 + 495 / 27 720 + 385 / 27 720 + 308 / 27 720 + 252 / 27 720 + 210 / 27 720 =
( 2 310 + 1 386 + 924 + 660 + 495 + 385 + 308 + 252 + 210 ) / 27 720 = 6 930 / 27 720
Đề thiếu chắc mk làm máy bài này rồi !
\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
= 1 + 2 + 2
= 5
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(2A=3A-A\)
\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(2A=\frac{3}{4}-\frac{1}{927}\)
\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)
\(A=\frac{182}{243}:\frac{1}{2}\)
\(A=\frac{364}{243}\)
\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};...;\frac{1}{20}=\frac{1}{20}\)
\(\Rightarrow A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{10.1}{20}\)
\(\Rightarrow A>\frac{1}{2}\)
Chúc bạn học tốt^^
\(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)
10 phân số
\(A>10.\frac{1}{20}=\frac{1}{2}\left(đpcm\right)\)