Chứng minh rằng
\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}>\frac{5}{6}\)
Chứng minh rằng
\(\frac{1}{5}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{44}+\frac{1}{45}>\frac{5}{6}\)
So sanh A va B:
\(A=\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}\)
\(B=\frac{5}{6}\)
a) \(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
b) \(\frac{\frac{15}{12}+\frac{3}{4}-1}{3-\frac{5}{6}+\frac{2}{3}}+\frac{\frac{16}{5}+\frac{16}{7}-\frac{16}{9}}{\frac{17}{5}+\frac{17}{7}-\frac{17}{9}}\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
Cho \(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}.\) .
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}.\)
Chứng minh rằng \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B.\)
bài này dài lắm
\(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}\)
\(A=\frac{\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)}{\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}\)
\(A=\frac{\left(\frac{1}{100}\right)}{\left(\frac{1}{25}\right)}=\frac{1}{4}\)
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}\)
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{6000}{2016}}{\frac{2000}{43}-\frac{2000}{2016}-\frac{2000}{257}}\)
\(B=\frac{16.\left(\frac{1}{9}-\frac{1}{127}+\frac{1}{2017}\right)}{5.\left(\frac{1}{2017}+\frac{1}{9}-\frac{1}{127}\right)}-\frac{6000.\left(\frac{1}{43}-\frac{1}{257}-\frac{1}{2016}\right)}{2000.\left(\frac{1}{43}-\frac{1}{2016}-\frac{1}{257}\right)}\)
\(B=\frac{16}{5}-3=\frac{1}{5}\)
Đặt \(C=\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}\)
\(C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2005^2}+\frac{1}{2006^2}+\frac{1}{2007^2}\)
\(C< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2004.2005}+\frac{1}{2005.2006}+\frac{1}{2006.2007}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2006}+\frac{1}{2006}-\frac{1}{2007}\)
\(=\frac{1}{4}-\frac{1}{2017}\left(đpcm\right)\)
\(C>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2005.2006}+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2007}-\frac{1}{2008}\)
\(=\frac{1}{5}-\frac{1}{2008}\left(đpcm\right)\)
Vậy \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B\)
Chứng minh rằng:\(\frac{43}{44}\le\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\le\frac{44}{45}\)
Chứng tỏ rằng: \(1< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+......+\frac{1}{16}+\frac{1}{17}< 2\)2
Chứng tỏ rằng: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}
1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 < 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 6/5 (1)
1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 < 1/11 + 1/11 + 1/11 + 1/11 +1/11 + 1/11 + 1/11 = 7/11 (2)
Từ (1) và (2) => :
A < 6/5 + 7/11 = 101/55 < 110/55 = 2
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1. Tính tích
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{899}{900}\)
2 Chứng tỏ rằng:\(y=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}<2\)
3. tính nhanh \(y=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
4. Chứng minh rằng \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}<1\)
Chứng tỏ rằng:
\(\frac{1}{15}\) +\(\frac{1}{16}\) +\(\frac{1}{17}\) +...+\(\frac{1}{44}\) >\(\frac{5}{6}\)
Số lượng phân số của dãy số trên là:
(44-15):1+1=30 (phân số)
Ta chia dãy phân số thành 2 cặp. Mỗi cặp có 15 phân số
Ta có: 1/15+1/16+1/17+...+1/44>5/6
Lại có: 1/30<1/15;1/30<1/16;...;1/30<1/29
1/45<1/30;1/45<1/31;...;1/45<1/44
=> 1/30.15+1/45.15 < 1/15+1/16+1/17+...+1/44
=> 15.(1/30+1/45)< 1/15+1/16+1/17+...+1/44
=> 15.1/18< 1/15+1/16+1/17+...+1/44
=> 5/6 < 1/15+1/16+1/17+...+1/44 (đpcm)
A> 1/29+1/29+......1/29+1/44+1/44.....+1/44
A> 15 x 1/29 + 15 x 1/44
Suy ra: (dựa vào tính chất hai phân số có cùng tử số phân số nào có mẫu số lớn hơn thì phân số đó nhỏ hơn)
A> 15 x 1/30 +15 x 1/45
A>1/2 +1/3
A> 5/6
Nhớ nhé
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