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\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=4.\frac{3200}{9603}=\frac{12800}{9603}\)
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\frac{3200}{9603}\)
\(A=\frac{800}{9603}\)
Bài trc mik làm lộn :)))
~ Hok tốt ~
A = 1/4.( 4/1.3.5 + 4/3.5.7+ ....+ 4/95.97.99)
= 1/4 .( 1/ 1.3 - 1/3.5 + 1/3.5 - 1/5.7 + .......+ 1/95.97 - 1/97.99)
= 1/4( 1/1.3 - 1/97.99)
= 1/4 . 9499/29397
a)\(A=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^2^5}\) <=>\(5A=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{24}}\)
<=>\(5A-A=(1+\frac{1}{5}+...+\frac{1}{5^{24}})-(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{25}})\)
<=>\(4A=1-\frac{1}{5^{25}}\) <=>\(A=\frac{(5^{25^{ }}-1)}{5^{25}}\div4\)
a, A = {abc; acb; bac; bca; cab; cba}
b, vì a < b < c
=> Số lớn nhất là cab
=> Số nhỏ nhất là abc
=> Theo đề bài, ta có:
cba + abc = 499
=> 100c + 10b + a + 100a + 10b + c = 499
=> 101c + 101a + 20b = 499
=> 101(c+a) + 20b = 499
Ờ đến đây tui tịt ùi. Chờ để tui nghĩ thêm
Đặt A = \(1.3.5+3.5.7+5.7.9+..+93.95.97+95.97.99\)
\(8A=1.3.5.8+3.5.7.8+...+93.95.97.8+95.97.99.8\)
\(8A=1.3.5.\left(1+7\right)+3.5.7.\left(9-1\right)+...+95.97.99\left(101-93\right)\)
\(8A=1.3.5.7+15+3.5.7.9-1.3.5.7+...+95.97.99.101-93.95.97.99\)
\(8A=15+95.97.99.101\)
\(\Rightarrow\) \(A=\frac{15+95.97.99.101}{8}\)
\(KL:........\)
Câu 1:a) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{11}+\frac{5}{12}\right)\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{7}{17}\)
\(=0+1+\frac{7}{17}\)
\(=\frac{17}{17}+\frac{7}{17}\)
\(=\frac{24}{17}\)
b) \(\frac{7}{12}-\left(\frac{5}{12}-\frac{5}{6}\right)\)
\(=\frac{7}{12}-\frac{5}{12}+\frac{5}{6}\)
\(=\frac{7}{12}-\frac{5}{12}+\frac{10}{12}\)
\(=\frac{7-5+10}{12}\)
\(=1\)
c) \(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{12}+\frac{1}{30}\)
\(=\frac{5}{60}+\frac{2}{60}\)
\(=\frac{7}{60}\)
Câu 2:a) \(\frac{x}{8}=2+\frac{-3}{2}\)
\(\Leftrightarrow\frac{x}{8}=\frac{4-3}{2}\)
\(\Leftrightarrow\frac{x}{8}=\frac{1}{2}\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=\frac{8}{2}\)
\(\Leftrightarrow x=4\)
b) \(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Leftrightarrow\frac{-18}{6}\le x\le4\)
\(\Leftrightarrow-3\le x\le4\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
B = \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+....+\frac{1}{95.97.99}\)
B = \(\frac{1}{4}.\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+...+\frac{99-95}{95.97.99}\right)\)
B = \(\frac{1}{4}.\left(\frac{5}{1.3.5}-\frac{1}{1.3.5}+\frac{7}{3.5.7}-\frac{3}{3.5.7}+...+\frac{99}{95.97.99}-\frac{95}{95.97.99}\right)\)
B = \(\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
B = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{97.99}\right)\)
B = \(\frac{1}{4}.\frac{3200}{9603}\)
B = \(\frac{800}{9603}\)