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Ta có A = \(\frac{10^{100}-1}{10^{98}-1}=\frac{10^{98}.10^2-10^2+99}{10^{98}-1}\)
\(=\frac{10^2\left(10^{98}-1\right)+99}{10^{98-1}}\)
\(=10^2+\frac{99}{10^{98}-1}\)
B= \(\frac{10^{101}-1}{10^{99}-1}=\frac{10^{99}.10^2-10^2+99}{10^{99}-1}\)
\(=\frac{10^2\left(10^{99}-1\right)+99}{10^{99}-1}\)
\(=10^2+\frac{99}{10^{99}-1}\)
Vì \(\frac{99}{10^{98}-1}>\frac{99}{10^{99}-1}\)nên \(10^2+\frac{99}{10^{98}-1}>10^2+\frac{99}{10^{99}-1}\)=> A > B
Vậy A > B
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2002}\)
<=>\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2002}+1\)
<=>\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
<=>\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
<=>\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
=>x+2004=0
<=>x=-2004
Vậy x=-2004
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Nhớ có lời giải nha các bạn , lm đc mk kết bạn với !!!! (^-^)
Ta chứng minh bài toán phụ:
Với a<b thì\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(c\inℕ^∗\right)\)
Ta có: \(a< b\)
\(\Rightarrow ac< bc\)
\(\Rightarrow ac+ba< bc+ba\)
\(a\left(b+c\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)
đpcm
Áp dụng vào bài toán ta có:
\(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}\)
Vậy \(\frac{10^{19}+1}{10^{20}+1}>\frac{10^{20}+1}{10^{21}+1}\)
Tham khảo nhé~
Đặt \(A=\frac{10^{19}+1}{10^{20}+1}\)
\(\Rightarrow10A=\frac{10^{20}+10}{10^{20}+1}=\frac{10^{20}+1+9}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
\(B=\frac{10^{20}+1}{10^{21}+1}\)
\(\Rightarrow10B=\frac{10^{21}+10}{10^{21}+1}=\frac{10^{21}+1+9}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
\(\Rightarrow\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\)
\(\Rightarrow1+\frac{9}{10^{20}+1}>1+\frac{9}{10^{21}+1}\)
\(\Rightarrow10A>10B\Rightarrow A>B\)
Ta có \(\frac{1}{9S}=\frac{9^{2017}+\frac{1}{9}}{9^{2017}+1}\)= \(\frac{9^{2017}+1-\frac{8}{9}}{9^{2017}+1}=1-\frac{\frac{8}{9}}{9^{2017}+1}\)
\(\frac{1}{9M}=\frac{9^{2016}+\frac{1}{9}}{9^{2016}+1}\)= \(\frac{9^{2016}+1-\frac{8}{9}}{9^{2016}+1}=1-\frac{\frac{8}{9}}{9^{2016}+1}\)
Vì \(9^{2016}+1< 9^{2017}+1\)=> \(\frac{\frac{8}{9}}{9^{2016}+1}>\frac{\frac{8}{9}}{9^{2017}+1}\)
=> \(1-\frac{\frac{8}{9}}{9^{2016}+1}< 1-\frac{\frac{8}{9}}{9^{2017}+1}\)=> \(\frac{1}{9}S< \frac{1}{9}M\Rightarrow S< M\)
ta có \(A=\frac{1}{100}+\frac{1}{101}+...+\frac{1}{149}\)
ta thấy \(\frac{1}{100}=\frac{1}{100}\)
\(\frac{1}{101}<\frac{1}{100}\)
\(\frac{1}{102}<\frac{1}{100}\)
................................
\(\frac{1}{149}<\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{149}<\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)
\(=\frac{49}{100}<\frac{1}{2}\)
vì \(A<\frac{49}{100}<\frac{1}{2}\Leftrightarrow A<\frac{1}{2}\)
Câu 1 :
Ta có : \(A=\frac{10^{100}+1}{10^{101}+1}\)
\(\Rightarrow10A=\frac{10^{101}+10}{10^{101}+1}=\frac{10^{101}+1+9}{10^{101}+1}=1+\frac{9}{10^{101}+1}\)
Ta có : \(B=\frac{10^{101}+1}{10^{102}+1}\)
\(10B=\frac{10^{102}+10}{10^{102}+1}=\frac{10^{102}+1+9}{10^{102}+1}=1+\frac{9}{10^{102}+1}\)
Vì 10101+1<10102+1
\(\Rightarrow\frac{9}{10^{101}+1}>\frac{9}{10^{102}+1}\)
\(\Rightarrow1+\frac{9}{10^{101}+1}>1+\frac{9}{10^{102}+1}\)
\(\Rightarrow\)10A>10B
\(\Rightarrow\)A>B
Vậy A>B.
Câu 2 :
Ta có : \(E=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì 2001<2001+2002 và 2002<2001+2002
\(\Rightarrow\hept{\begin{cases}\frac{2000}{2001}>\frac{2000}{2001+2002}\\\frac{2001}{2002}>\frac{2001}{2001+2002}\end{cases}}\)
\(\Rightarrow C>E\)
Vậy C>E.