Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tớ không chép lại đề nữa nhé:
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)=\(\frac{1}{2}.\left(\frac{3-1}{1-3}+\frac{7-5}{5-7}+...+\frac{2011-2009}{2009-2011}\right)\)
= \(\frac{1}{2}.\left(\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+...+\frac{2011}{2009.2011}-\frac{2009}{2009.2011}\right)\)
=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
=\(\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
=\(\frac{1}{2}.\frac{2010}{2011}\)
=\(\frac{1005}{2011}\)
Ta có: \(\frac{1999x2000}{1999x2000+1}=\frac{1999x2000+1-1}{1999x2000+1}=1-\frac{1}{1999x2000+1}\)
\(\frac{2000x2001}{2000x2001+1}=\frac{2000x2001+1-1}{2000x2001+1}=1-\frac{1}{2000x2001+1}\)
Nhận thấy: \(\frac{1}{1999x2000+1}>\frac{1}{2000x2001+1}\)=> \(1-\frac{1}{1999x2000+1}< 1-\frac{1}{2000x2001+1}\)
=> \(\frac{1999x2000}{1999x2000+1}=\frac{2000x2001}{2000x2001+1}\)
\(\frac{1999x2000}{1999x2000+1}< \frac{2000x2001}{2000x2001+1}\)
Ta thấy 36 là BCNN( 18, 12, 9, 4) nên ta có:
\(\frac{1}{18}=\frac{1\times2}{18\times2}=\frac{2}{36};\frac{x}{12}=\frac{x\times3}{12\times3}=\frac{x\times3}{36};\frac{y}{9}=\frac{y\times4}{9\times4}=\frac{y\times4}{36};\frac{1}{4}=\frac{1\times9}{4\times9}=\frac{9}{36}\)\(=\frac{9}{36}\)
quy đồng xong ta có
\(\frac{2}{36}< \frac{x\times3}{36}< \frac{y\times4}{36}< \frac{9}{36}\)
để thoã mãn điều kiện trên vậy x=1;y=2
\(C=\frac{1999^{2000}+1}{1999^{1999}+1}< \frac{1999^{1999}+1+1998}{1999^{2000}+1+1998}\)
\(=\frac{1999^{1999}+1999}{1999^{2000}+1999}\)
\(=\frac{1999\cdot(1999^{1998}+1)}{1999\cdot(1999^{1999}+1)}\)
\(=\frac{1999^{1999}+1}{1999^{1998}+1}=D\)
Vậy...
Ta có
\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}};\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}\)
\(\Rightarrow\frac{1}{3^4}< \frac{1}{4^3}\left(3^4>4^3\right)\\
\Rightarrow\frac{1}{3^{400}}< \frac{1}{4^{300}}\)
\(\frac{-1999}{2000}\)>\(\frac{-1999}{2001}\)>\(\frac{-2000}{2001}\)
Ta có:
\(\frac{19992000}{20002000}=\frac{19991999+1}{20002000}=\frac{19991999}{20002000}+\frac{1}{20002000}\)
\(=\frac{1999}{2000}+\frac{1}{20002000}\)
Vì \(\frac{1999}{2000}< \frac{1999}{2000}+\frac{1}{20002000}\Rightarrow\frac{1999}{2000}< \frac{19992002}{20002000}\)
Ta có : \(\frac{19992000}{20002000}\)\(=\)\(\frac{19991999+1}{20002000}\)\(=\)\(\frac{19991999}{20002000}+\frac{1}{20002000}\)
\(=\frac{1999}{2000}+\frac{1}{20002000}\)
Vì \(\frac{1999}{2000}< \frac{1999}{2000}+\frac{1}{20002000}=\frac{1999}{2000}< \frac{19992002}{20002000}\)