A= \(\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{\eta.\left(\eta+30\right)}+...+\frac{1}{1973.2003}\)
B= \(\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{\eta.\left(\eta+1972\right)}+...+\frac{1}{31.2003}\)
Những câu hỏi liên quan
Tính \(\frac{A}{B}\)biết :
\(A=\frac{1}{2.32}+\frac{1}{3.33}+....+\frac{1}{n\left(n+30\right)}+....+\frac{1}{1973.2003}\)
\(B=\frac{1}{2.1974}+\frac{1}{3.1975}+....+\frac{1}{n\left(n+1972\right)}+....+\frac{1}{31.2003}\)
\(A=\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{1973.2003}\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}-\frac{1}{32}-\frac{1}{33}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
\(B=\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{31.2003}\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
Vậy \(\frac{A}{B}=\frac{1972}{30}\)
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cho: \(\frac{\alpha}{b}=\frac{b}{c}=\frac{c}{\eta}.cmr\left(\frac{\alpha+b+c}{b+c+\eta}\right)^3=\frac{\alpha}{\eta}\)
1. Tìm m sao cho \(y=\frac{m\sin x+4}{\sin x+m}\)nghịch biến trên \(\left(0,\frac{\eta}{2}\right)\)
2. Tìm m sao cho \(y=\frac{\cos x+1}{m\cos x+2}\)nghịch biến trên \(\left(0,\frac{\eta}{2}\right)\)
rút gọn
\(C=\frac{\sqrt{2}\cos x-2\cos\left(\frac{\eta}{4}+x\right)}{-\sqrt{2}\sin x-2\sin\left(\frac{\eta}{4}+x\right)}\)
Everybody =) Help me.
Let be given an open interval\(\left(\alpha;eta\right)\) with \(eta-\alpha=\frac{1}{2007}.\)Determine the maximum number of irreducible fractions \(\frac{a}{b}\) in \(\left(\alpha;eta\right)\) with \(1\le b\le2007?\)
A) 1002
b) 1003
C) 1004
D)1005
E)1006
mình rất muốn giúp bạn nhưng cuộc sống đẹp ở chỗ là mình học ngu anh văn :v, nên mình xin lỗi
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\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\).Tìm a, b, c, d \(\eta\)
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a=1;b=2;c=3;d=4
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Tính P/Q biết:
P = 1/2.32 + 1/3.33 + ... + 1/n.(n+30) + ... + 1/1973.2003
Q = 1/2.1974 + 1/3.1975 + ... + 1/n.(n+1972) + ... + 1/31.2003
\(P=...\)
\(=\frac{1}{30}\left(\frac{30}{2.32}+\frac{30}{3.33}+...+\frac{30}{1973.2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}\right)-\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{2003}\right)\right]\)
\(=\frac{1}{30}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]\)
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\(Q=...\)
\(=\frac{1}{1972}\left(\frac{1972}{2.1974}+\frac{1972}{3.1975}+...+\frac{1}{31.2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]\)
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Gọi \(\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]=A\)
Ta có:\(\frac{P}{Q}=\left(\frac{1}{30}.A\right):\left(\frac{1}{1972}.A\right)=\frac{A}{30}\cdot\frac{1972}{A}=\frac{1972}{30}=\frac{986}{15}\)
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Tính \(\frac{A}{B}\)biết:
\(A=\frac{1}{2.32}+\frac{1}{3.33}+\frac{1}{4.34}+...+\frac{1}{1973.2003}\)
\(B=\frac{1}{2.1074}+\frac{1}{3.1075}+\frac{1}{4.1076}+...+\frac{1}{34.2003}\)
Tính:\(\frac{2\eta-2}{\eta-3}\)
Ta có: \(\frac{2n-2}{n-3}=\frac{2n-6+4}{n-3}\)
\(=\frac{2\left(n-3\right)+4}{n-3}=\frac{2\left(n-3\right)}{n-3}+\frac{4}{n-3}\)
\(=2+\frac{4}{n-3}\left(n\ne3\right)\)
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ta co : \(\frac{2n-2}{n-3}=\frac{2\left(n-3\right)+4}{n-3}=\frac{2\left(n-3\right)}{n-3}+\frac{4}{n-3}=2+\frac{4}{n-3}\left(n\ne3\right)\)
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