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BÀi 1
Vì \(a+m\ge a\)
\(b+m\ge b\)
\(\Rightarrow\frac{a+m}{b+m}< \frac{a}{b}\)
hok tốt
bài 1 ngắn vậy à?
ai làm bài 2 giúp mình đi
mình cần gấp, 2 hôm nữa phải nộp rồi
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\(B-A=\frac{11-10}{a^m}+\frac{9-10}{a^n}=\frac{1}{a^m}-\frac{1}{a^n}\)
Nếu \(m>n\) thì \(\frac{1}{a^m}-\frac{1}{a^n}< 0\Rightarrow B< A\)
Nếu \(m< n\) thì \(\frac{1}{a^m}-\frac{1}{a^n}>0\Rightarrow B>A\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
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.
A=\(\frac{-199}{10^{2011}}\)
B=\(\frac{-109}{10^{2011}}\)
Dễ dàng so sánh được A<B
Ta có : \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1000.1001}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1001-1000}{1000.1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}=\frac{1000}{1001}\)
Ta thấy : \(1001< 2020\Rightarrow\frac{1}{1001}>\frac{1}{2020}\)
\(\Rightarrow-\frac{1}{1001}< -\frac{1}{2020}\)
\(\Rightarrow1-\frac{1}{1001}< 1-\frac{1}{2020}\Rightarrow\frac{1000}{1001}< \frac{2019}{2020}\)
Hay : \(N< M\)
Coi \(A=\frac{10}{a^m}+\frac{10}{a^n}=\frac{9}{a^m}+\frac{10}{a^n}+\frac{1}{a^m}\)
\(B=\frac{9}{a^m}+\frac{11}{a^n}=\frac{9}{a^m}+\frac{10}{a^n}+\frac{1}{a^n}\)
Cả A và B đều có: \(\frac{9}{a^m}+\frac{10}{a^n}\) nên ta so sánh \(\frac{1}{a^n}\)và\(\frac{1}{a^m}\)
TH1: n<m =>1/n>1/m
=>B>A
TH2:n>m=>1/n<1/m
=>B<A
TH3: m=n =>1/m=1/n
=> B=A
\(\frac{10}{a^m}+\frac{10}{a^n}=\left(\frac{9}{a^m}+\frac{10}{a^n}\right)+\frac{1}{a^m}\)
\(\frac{9}{a^m}+\frac{11}{a^n}=\left(\frac{9}{a^m}+\frac{10}{a^n}\right)+\frac{1}{a^n}\)
Muốn so sách 2 biểu thức trên ta chỉ cần so sánh \(\frac{1}{a^m}\) với \(\frac{1}{a^n}\)
Trường hợp 1: a=1 thì 2 biểu thức trên = nhau
Trường hợp 2: a khác 1 thì xét m và n
-Nếu m=n thì am=an => 2 biểu thức trên = nhau
-Nếu m<n thì am<an => \(\frac{1}{a^m}>\frac{1}{a^n}\)=> .....
-Nếu m>N thì am>an => \(\frac{1}{a^m}<\frac{1}{a^n}\)=> ......