Cho A = 1/2^2 + 1/2^3 +...............+1/9^2
CMR 8/9>A>2/5
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Cho A=1/2^2+1/3^2+1/4^2+......+1/9^2
CMR 8/9 lớn hơn A lớn hơn 2/5
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
Tương tự:\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\left(2\right)\)
Từ (1) và (2) => \(\frac{8}{9}>A>\frac{2}{5}\left(đpcm\right)\)
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CMR : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Với \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)
\(\dfrac{1}{1\cdot2}>\dfrac{1}{2^2}>\dfrac{1}{2\cdot3},\dfrac{1}{2\cdot3}>\dfrac{1}{3^2}>\dfrac{1}{3\cdot4},...,\dfrac{1}{8\cdot9}>\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(\Rightarrow1-\dfrac{1}{9}>A>\dfrac{1}{2}-\dfrac{1}{10}\) \(\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\)
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CMR 2/5<A<8/9 với A=1/2^2+1/3^2+...+1/9^2
làm giúp mik
Các bạn giúp mình nhé , mk đang cần gấp , bạn nào giải hộ , mk sẽ tick đều đặn. Help me1.Cmr : A9/10!+9/11!+9/12!+...+9/1000! 1/92. CHo G 5/3+8/3^2+11/3^3+...+302/3^100. CMR : 23/9G7/23.so sánh : L (1-1/2)(1-1/3)...(1-1/20) với 1/214.C1/101+1/102+...+1/200. CMR:a/ C7/12b//C5/85 cho C 1/11+1/12+...+1/13+...+1/70CMR : 4/3C2,56. Cho B 4/3+10/9+28/27+...+399/398 . CMR B 100
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Các bạn giúp mình nhé , mk đang cần gấp , bạn nào giải hộ , mk sẽ tick đều đặn. Help me
1.Cmr : A=9/10!+9/11!+9/12!+...+9/1000! < 1/9
2. CHo G = 5/3+8/3^2+11/3^3+...+302/3^100. CMR : 23/9<G<7/2
3.so sánh : L =(1-1/2)(1-1/3)...(1-1/20) với 1/21
4.C=1/101+1/102+...+1/200. CMR:
a/ C>7/12
b//C>5/8
5 cho C = 1/11+1/12+...+1/13+...+1/70
CMR : 4/3<C<2,5
6. Cho B = 4/3+10/9+28/27+...+399/398 . CMR B< 100
cho P =\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}cmr\dfrac{2}{5}< P< \dfrac{8}{9}\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
\(P=\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{9.9}\)
\(P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(P=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(P=1-\dfrac{1}{9}=\dfrac{8}{9}\)
\(\Rightarrow P< \dfrac{8}{9}\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
\(P=\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{9.9}\)
\(P>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(P=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{9}-\dfrac{1}{10}\)
\(P=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
\(\Rightarrow P>\dfrac{2}{5}\)
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1.8,cho A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\).CMR:\(\dfrac{2}{5}< A< \dfrac{8}{9}\)
1.9,cho A=\(\dfrac{2}{3}+\dfrac{2}{5^2}+\dfrac{2}{7^2}+...+\dfrac{2}{2007^2}.CMR:A< \dfrac{1007}{2008}\)
Câu 1.8: Giải
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(A>\dfrac{2}{5}\) (1)
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(A< 1-\dfrac{1}{9}\)
\(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
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cho s =1/22+1/3^2+1/4^2+...........+1/9^2. cmr:2/5 < s < 8/9
Lời giải:
$S=\frac{1}{2^2}+\frac{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}$
$> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}$
$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}$
$=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}(*)$
Lại có:
$S=\frac{1}{2^2}+\frac{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}(**)$
Từ $(*); (**)$ ta có đpcm.
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Cho tong S = 1/2^2 + 1/3^2 +1/4^2 +.....+ 1/9^2
CMR : 2/5 < S <8/9
a)Cho A= 1/2^2+1/3^2+...+1/n^2.CMR A<1
b)Cho B=1/2^2+1/4^2+1/6^2+...+1/(2n)^2.CMR B<1/2
c)Cho C=3/4+8/9+15/16+...+n^2-1/n^2.CMR C<n-2