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a) \(\frac{3}{-4}=\frac{-3}{4};\frac{-1}{-4}=\frac{1}{4}\)
Vì - 3 < 1 nên \(\frac{-3}{4}< \frac{1}{4}\)
hay \(\frac{3}{-4}< \frac{-1}{-4}\)
Quy đồng mẫu ta được:
15/17=15.27/17.27=405/459
25/27=25.17/27.27=425/459
⇒405/459<425/459⇒15/17<25/27
Ta có:
\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(\frac{1}{10^{16}+1}>\frac{1}{10^{17}+1}\)
\(\Rightarrow\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\)
\(\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
\(\Rightarrow A>B\)
a) Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
\(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(\frac{1}{10}C=\frac{10^{1992}+1}{10^{1992}+10}=1+\frac{10^{1992}+1}{9}\)
\(\frac{1}{10}D=\frac{10^{1993}+1}{10^{1993}+10}=1+\frac{10^{1993}+1}{9}\)
\(\frac{10^{1992}+1}{9}< \frac{10^{1993}+1}{9}\Rightarrow1+\frac{10^{1992}+1}{9}< 1+\frac{10^{1993}+1}{9}\)
\(\Rightarrow\frac{1}{10}C< \frac{1}{10}D\)
\(\Rightarrow C< D\)
Vậy C < D
đặt A=\(\frac{10^{2011}+10}{10^{2012}+10}\)
=>10A=\(\frac{10\left(10^{2011}+10\right)}{10^{2012}+10}=\frac{10^{2012}+100}{10^{2012}+10}=\frac{10^{2012}+10}{10^{2012}+10}+\frac{90}{10^{2012}+10}=1+\frac{90}{10^{2012}+10}\)
đặt B=\(\frac{10^{2012}-10}{10^{2013}-10}\)
=>10B=\(\frac{10\left(10^{2012}-10\right)}{10^{2013}-10}=\frac{10^{2013}-100}{10^{2013}-10}=\frac{10^{2013}-10}{10^{2013}-10}+\frac{-90}{10^{2013}-10}=1+\frac{-90}{10^{2013}-10}\)
vì \(\frac{-90}{10^{2013}-10}\) luôn âm nên
\(1+\frac{90}{10^{2012}+10}>1+\frac{-90}{10^{2013}-10}\)
vậy \(A>Bhay\frac{10^{2011}+10}{10^{2012}+10}>\frac{10^{2012}-10}{10^{2013}-10}\)
ta có A=\(\frac{17^{18}+1}{17^{19}+1}\)<\(\frac{17^{18}+1+16}{17^{19}+1+16}\) (nếu a/b<1 thì a+c/b+c>a/b)
A<\(\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}\)
A,<\(\frac{17^{17}+1}{17^{18}+1}\)=B
hay A<B
\(A=\frac{17^{18}+1}{17^{19}+1}\) với \(B=\frac{17^{17}+1}{17^{18}+1}\)
Ta có :B=\(\frac{17^{17}+1}{17^{18}+1}=\frac{17^{18}+17}{17^{19}+17}\)
Ta có:1-B=\(1-\frac{17^{18}+17}{17^{19}+17}=\frac{17^{19}+17-17^{18}-17}{17^{19}+17}=\frac{17^{19}-17^{18}}{17^{19}+17}\)
1-A=1-\(\frac{17^{18}+1}{17^{19}+1}=\frac{17^{19}+1-17^{18}-1}{17^{19}+1}=\frac{17^{19}-17^{18}}{17^{19}+1}\)
Do \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)
\(\Rightarrow A< B\)
\(\frac{429}{639}>\frac{219}{333}\)
\(\frac{10^{15}+1}{10^{16}+1}>\frac{10^{16}+1}{10^{17}+1}\)
ngu quá