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B=(10101+1):(10102+1)<(10101+1+9):(10102 +1+9)=(10101+10):(10102+10)=[10.(10100+1]:[10.(10101+)]
=(10100+1):(10101+1)=A
=>A>B
vì 2017100 + 1 < 2017101 + 1
\(\Rightarrow\frac{2017^{100}+1}{2017^{101}+1}< \frac{2017^{100}+1+2016}{2017^{101}+1+2016}=\frac{2017^{100}+2017}{2017^{101}+2017}=\frac{2017.\left(2017^{99+1}\right)}{2017.\left(2017^{100}+1\right)}=\frac{2017^{99}+1}{2017^{100}+1}\)
Vậy \(\frac{2017^{99}+1}{2017^{100}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
so sánh 2 phân số cùng mẫu thì ta xét tử
đừng nói không làm được chứ
Ta có: \(A=\frac{2017^{99}+1}{2017^{100}+1}\Rightarrow2017A=\frac{2017^{100}+2017}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\Rightarrow2017B=\frac{2017^{101}+2017}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
\(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
\(\Rightarrow2017A>2017B\Rightarrow A>B\)
Vậy...
Đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\)nên \(2017A=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\)nên \(2017B=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
Vì \(1=1;\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
Hay \(2017A>2017B\)nên \(A>B\)
Vây \(\frac{2017^{99}+1}{2017^{1001}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
A=\(\frac{10^{2015}+1}{10^{2016}+1}\)=>10A=\(\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}\)= \(\frac{10^{2016}+10}{10^{2016}+1}\)=\(\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}\)=\(\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}\)=1+\(\frac{9}{10^{2016}+1}\)
B=\(\frac{10^{2016}+1}{10^{2017}+1}\)=>10B=\(\frac{10.\left(10^{2016}+1\right)}{10^{2017+1}}=\frac{10^{2017}+10}{10^{2017}+1}\)= \(\frac{\left(10^{2017}+1\right)+9}{10^{2017}+1}\)=\(\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}\)= 1+\(\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{17}+1\)=>\(\frac{9}{10^{2016}+1}\)>\(\frac{9}{10^{2017}+1}\)nên \(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)=>10A>10B
Vậy A>B
Ta có: 100+101/101+102
= 100/101+102 + 101/101+102
Vì 100/101>100/101+102
101/102 > 101/101+102
=>100/101+101/102 > 100+101/101+102
\(a,\frac{8}{9}< \frac{108}{109}\)
\(b,\frac{97}{100}< \frac{98}{99}\)
\(c,\frac{19}{18}>\frac{2017}{2016}\)
\(d,\frac{15}{16}>\frac{515}{616}\)
vì -2016/2017<1 và 1/10>1 => -2016/2017<1/10
Ta có 99/-100=-99/100>-1 nên 99/-100>1 và vì -102/101<-1 =>99/-100>-102/101