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
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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A= \(\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{\eta.\left(\eta+30\right)}+...+\frac{1}{1973.2003}\)
B= \(\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{\eta.\left(\eta+1972\right)}+...+\frac{1}{31.2003}\)
Tính tỉ số A/B biết: A=1/22.32+1/3.3+...+1/1073.2003
B=1/2.1974+1/3.1975+...+1/31.2003
vào câu hỏi tương tự
Câu tìm câu hỏi tương tự mà nghĩ ra
Ta có : \(A=\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{1973.2003}\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+....+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
Lại có : \(B=\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{31.2003}\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+....+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
Khi đó : \(\frac{A}{B}=\frac{1972}{30}\)
Xem thêm câu trả lời
Tính A, biết: (tính nhanh, ko tính máy tính)
\(A=\frac{1}{2.32}+\frac{1}{3.33}+\frac{1}{4.34}\)
30A=30/2*32+30/3*33+30/4*34=1/2-1/32+1/3-1/33+1/4-1/34=99/100
A=3,3/100
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\(=\frac{1}{30}.\left(\frac{30}{2.32}+\frac{30}{3.33}+\frac{30}{4.34}\right)\)
\(=\frac{1}{30}.\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+\frac{1}{4}-\frac{1}{34}\right)\)
Tự làm tiếp nhé
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R=\(\dfrac{1}{2.32}+\dfrac{1}{3.33}+...+\dfrac{1}{1976+2006}\)
S=\(\dfrac{1}{2.1977}+\dfrac{1}{3.1978}+...+\dfrac{1}{31.2006}\)
tinh \(\dfrac{R}{S}\)
\(R=\frac{1}{2.32}+\frac{1}{3.33}+......+\frac{1}{1976.2006}\Rightarrow30R=\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1976}-\frac{1}{32}-\frac{1}{33}-....-\frac{1}{2006}=\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{31}-\frac{1}{1977}-\frac{1}{1978}-....-\frac{1}{2006};S=\frac{1}{2.1977}+\frac{1}{3.1978}+....+\frac{1}{31.2006}=\Rightarrow1975S=\frac{1}{2}+\frac{1}{3}+....+\frac{1}{31}-\frac{1}{1977}-\frac{1}{1978}-....-\frac{1}{2006}=R\Rightarrow30R=1975S\Rightarrow R=\frac{1975}{30}S=\frac{395}{6}\Rightarrow\frac{R}{S}=\frac{395}{6}\)
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Tính \(\frac{A}{B}\)biết:
\(A=\frac{1}{2.32}+\frac{1}{3.33}+\frac{1}{4.34}+...+\frac{1}{1973.2003}\)
\(B=\frac{1}{2.1074}+\frac{1}{3.1075}+\frac{1}{4.1076}+...+\frac{1}{34.2003}\)
tinh A/B, biet
A=1/2*32+1/3*33+1/4*34+...+1/n*(n+30)+...+1/1973*2003
B=1/2*1974+1/3*1975+1/4*1976+...+1/n*(n+1972)+...+1/31*2003.
tìm n biết:p=(n-2).(n2+n-1)là số nguyên tố
chứng tỏ với mọi n\(\in\)N* ta luôn có:\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
áp dụng tính tổng sau:\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
chứng tỏ :
Ta có : \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
áp dụng :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=1-\frac{1}{9}\)
\(A=\frac{8}{9}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=1-\frac{1}{9}=\frac{8}{9}\)
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