không tính, hãy chứng tỏ : S = \(\frac{29}{21}\) + \(\frac{16}{23}\) + \(\frac{5}{8}\) < 3
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29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29 = -5
1 + 29-x/21 + 1 + 27-x/23 + 1 + 25-x/25 + 1 + 23-x/27 + 1 + 21-x/29 = 0
50-x/21 + 50-x/23 + 50-x/25 + 50-x/27 + 50-x/29 = 0
(50-x) (1/21 + 1/23 + 1/25 + 1/27 + 1/29) = 0
Vì: 1/21 + 1/23 + 1/25 + 1/27 + 1/2 > 0
=> 50 - x = 0
x = 50
Vậy x = 50
\(\frac{-1}{3}+\frac{0,2-0,3+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{10}+\frac{5}{11}}{\frac{-3}{10}+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{39}{110}}{\frac{-79}{80}}\)
\(=\frac{-1}{3}-\frac{312}{869}\)
\(=\frac{-1805}{2607}\)
Ta có \(S=5.\left(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\right)\)
\(S>5.\left(\frac{1}{49}+\frac{1}{49}+...+\frac{1}{49}\right)\)30 số hạng
\(S>5.\frac{30}{49}\)
\(S>\frac{150}{49}\)
\(S>3\frac{3}{49}\)
Suy ra \(S
Bài 1:
a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)
\(=-\frac{891}{25}.\frac{1}{4}\)
\(=-\frac{891}{100}\)
b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)
\(=\frac{3}{8}.\left(-14\right)\)
\(=-\frac{21}{4}\)
c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)
\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(=1+1=2\)
d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
Đặt \(A=\frac{9+\frac{9}{11}+\frac{18}{23}-\frac{27}{37}}{8+\frac{8}{11}+\frac{16}{23}-\frac{24}{37}}-\frac{2+\frac{16}{29}-\frac{24}{13}-\frac{32}{11}}{3+\frac{24}{29}-\frac{36}{13}-\frac{48}{11}}\)\(=\frac{9\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}{8\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}-\frac{2\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}{3\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}\)
\(=\frac{9}{8}-\frac{2}{3}\)(do \(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37};1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\ne0\))
\(=\frac{27}{24}-\frac{16}{24}=\frac{11}{24}.\)
Vậy A = \(\frac{11}{24}.\)
Ta có:
1/2 + 1/3 + 1/4 + ... + 1/15 + 1/16 = (1/2 + 1/3 + 1/4 + 1/5) + (1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11) + (1/12 + 1/13 + 1/14) + (1/15 + 1/16)
Vì 1/6 + 1/7 + 1/8 < 3x 1/6 = 1/2
1/9 + 1/10 + 1/11 <3x1/9 = 1/3
1/12 + 1/13 +1/14 < 3x1/12 = 1/4
1/15 + 1/16 < 3 x 1/15 = 1/5
Nên A < 2 x (1/2 + 1/3 + 1/4 + 1/5) < 2 x (1/2 + 1/2 + 1/4 + 1/4) =3 (1)
Lập luận tương tự có:
A = ( 1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12) + (1/13 + 1/14 + 1/15 + 1/16) > (1/2 + 1/3 + 1/4) + 4 x 1/8 + 4 x 1/ 12 + 4 x 1/16
Hay A > 2 x (1/2 + 1/3 + 1/4) > 2 x (1/2 + 1/4 + 1/4) = 2 (2)
Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.