1944-1945+1946+1947
Cảm ơn
Những câu hỏi liên quan
1/1945*1945+1/1946*1946+1/1947*1947+...+1/1974+*1974+1/1975*1975<1/1944
chung minh
1/1945^2+1/1946^2+1/1947^2+...+1/1975<1/1944
1/19452 < 1/ 1944.1945
1/19462 < 1/ 1945.1946
....
1/19752 < 1/ 1974.1975
=> 1/119452 +1/119462+....+1/119752 < 1/ 1944.1945+1/ 1945.1946+..+1/ 1974.1975=1/1944-1/1945+1/1945-1/1946+....+1/1974-1/1975
=1/19444-1/1975<1/1944
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Chứng tỏ rằng:
1/1945² + 1/1946² + 1/1947² + .......+1/1974² + 1/1975² < 1/1944
Chứng tỏ : 1/1945^2+1/1946^3+1/1947^2+...+1/1975^2<1/1944
\(\dfrac{1}{1945^2}< \dfrac{1}{1944^2}\\ \dfrac{1}{1946^2}< \dfrac{1}{1944^2}\\ \dfrac{1}{1947^2}< \dfrac{1}{1944^2}\\ ...\\ \dfrac{1}{1975^2}< \dfrac{1}{1944^2}\\ \Leftrightarrow\dfrac{1}{1945^2}+\dfrac{1}{1946^2}+\dfrac{1}{1947^2}+...+\dfrac{1}{1975^2}< \dfrac{1}{1944^2}+\dfrac{1}{1944^2}+\dfrac{1}{1944^2}+...+\dfrac{1}{1944^2}\left(31\text{ số }\dfrac{1}{1944^2}\right)=31\cdot\dfrac{1}{1944^2}< 1944\cdot\dfrac{1}{1944^2}=\dfrac{1}{1944}\)
Vậy \(\dfrac{1}{1945^2}+\dfrac{1}{1946^2}+\dfrac{1}{1947^2}+...+\dfrac{1}{1975^2}< \dfrac{1}{1944}\)
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Chứng tỏ rằng:
1/1945^2+1/1946^2+...+1/1947^2+1/1975^2<1/1944
Có : 1/1945^2 + 1/1946^2 + ...... + 1/1975^2
< 1/1944.1945 + 1/1945.1946 + ...... + 1/1974.1975
= 1/1944 - 1/1945 +1/1945 - 1/1946 + ...... + 1/1974 - 1/1975
= 1/1944 - 1/1975
< 1/1944
Tk mk nha
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Chứng tỏ rằng 1/1945^2+1?1946^2+...+1/1975^2<1/1944. Ai pít thì giúp mình ná
m.n ơi cho em hỏi zới !!!!
Chứng minh răng :
1/1945^2+1/1946^2+1/1947^2+.....+1/1974^2+1/1975^2<1/1944
Ta có \(\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1975^2}\)
\(< \frac{1}{1944\cdot1945}+\frac{1}{1945\cdot1946}+...+\frac{1}{1974.1975}\)
\(=\frac{1}{1944}-\frac{1}{1945}+\frac{1}{1945}-\frac{1}{1946}+...+\frac{1}{1974}-\frac{1}{1975}\)
=\(\frac{1}{1944}-\frac{1}{1975}< \frac{1}{1944}\)
\(\Rightarrow\frac{1}{1945^2}+\frac{1}{1946^2}+\frac{1}{1947^2}+..+\frac{1}{1975^2}< \frac{1}{1944}\)
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Chứng tỏ rằng
\(\frac{1}{1945^2}\)\(+\frac{1}{1946^2}+\frac{1}{1947^2}+...+\frac{1}{1974^2}+\frac{1}{1975^2}\)<1/1944
Xem thêm câu trả lời
Cmr: A=11^3+12^3+13^3+..+1944^3+1945^3 chia hết cho 6
\(A=11^3+12^3+...+1945^3\)
Ta có: \(A=11^3+12^3+...+1945^3\)
\(=\left(12^3+14^3+...+1944^3\right)+\left(11^3+13^3+...+1945^3\right)\)
Do dãy \(11;13;...;1945\) có \(\frac{1945-11}{2}+1=968\) số hạng
\(\Rightarrow \left(11^3+13^3+...+1945^3\right)⋮2\) mà \(\left(12^3+14^3+...+1944^3\right)⋮2\)
\(\Rightarrow A⋮2\left(1\right)\)
Mặt khác:
\(A=\left(11^3+1945^3\right)+\left(12^3+1944^3\right)+...+\left(977^3+979^3\right)+978^3\)
\(=967.1956^3+978^3⋮3\)
\(\Rightarrow A⋮3\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow A⋮6\)