chứng tỏ: 3/5.11 + 5/11.21 + 7/21.35 + 9/35.53 < 1
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3/5.11+5/11.21+7/21.35+9/35.53
3/5.11+3/11.21+7/21.35+9/35.53
CMR 3/5.11+5/11.21+7/21.35+9/35.53<1
Giải Nhanh Hộ Mình Cái
Chứng tỏ rằng A là số chính phương biết A=1+3+5+7+9+11+....+(2n-1)
1/Chứng tỏ 77 là ước của A=76+75-74
2/Cho A=2+22+23+...+260.Chứng tỏ rằng A là bội của 3, của 7 và của 15
3/Cho B=1+5+52+53+...+596+597+598. Chứng tỏ B chia hết cho 31
Chứng tỏ rằng: 1/3^3 + 1/5^3 + 1/7^3 +..+ 1/2021^3 < 1/12
Ta có \(\dfrac{1}{3^3}< \dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
\(\dfrac{1}{5^3}< \dfrac{1}{4.5.6}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)\\ ...\\ \dfrac{1}{2021^3}< \dfrac{1}{2020.2021.2022}=\dfrac{1}{2}\left(\dfrac{1}{2020.2021}-\dfrac{1}{2021.2022}\right)\)
Cộng VTV ta được
\(VT< \dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2021.2022}\right)=\dfrac{1}{12}-\dfrac{1}{2\left(2021.2022\right)}< \dfrac{1}{12}\)
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\(n^3=n.n^2>n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}.\dfrac{n+1-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)
\(\dfrac{1}{3^3}+\dfrac{1}{5^3}+.......+\dfrac{1}{2009^3}< \dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+.....\dfrac{1}{2008.2009.2010}=\dfrac{1}{2}\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+.........+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2009.2010}\right)\)
\(=\dfrac{1}{2}\)
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Biết rằng A=717 + 17. 3 - 1 chia hết cho 9. Chứng tỏ B = 718 + 18.3-1 chia hết cho 9
73=343 đồng dư với 1(mod 9)
=>(73)6=718 đồng dư với 1(mod 9)
=>718=9k+1
=>B=9k+1+18.3-1=9k+18.3=9(k+2.3) chia hết cho 9
=>đpcm
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Chứng tỏ rằng
1/2×3/4×5/6×7/8×……×99/100<1/10
Chứng tỏ:
1/26+1/27+...+1/49+1/50=99/50-97/49+...+7/4-5/3+3/2-1
Xét vế phải :
\(VT=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(\text{Nhầm xíu , cho sửa lại nhé}\)
\(\text{Xét vế phải :}\)
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)