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\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}\) theo công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
\(A< \frac{10^{11}+10}{10^{12}+10}=\frac{10^{10}\left(10+1\right)}{10^{11}\left(10+1\right)}=\frac{10^{10}}{10^{11}}\)
\(\Rightarrow\frac{10^{10}}{10^{11}}=\frac{10^{10}\cdot10^{12}}{10^{11}\cdot10^{12}}=\frac{10^{22}}{10^{23}}\)
\(\Leftrightarrow A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}\)
Lại áp dụng công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
\(A< \frac{10^{10}}{10^{11}}=\frac{10^{11}}{10^{12}}< \frac{10^{11}+1}{10^{12}+1}=B\)
\(\Leftrightarrow A< B\)
Hoặc \(A< \frac{10^{11}-1+2}{10^{12}-1+2}=\frac{10^{12}+1}{10^{12}+1}\)
..... (EZ)
Ta có :
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
\(\Rightarrow A< B\)
So sánh 2 phân số sau $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10
kick dzô chữ xanh là được!! OK
Ta có :
10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)
= \(\frac{10^{2012}+10}{10^{2012}+1}\)
= \(\frac{10^{2012}+1+9}{10^{2012}+1}\)
= \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)
= 1 - \(\frac{9}{10^{2012}+1}\)
10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)
= \(\frac{10^{2013}+10}{10^{2013}+1}\)
= \(\frac{10^{2013}+1+9}{10^{2013}+1}\)
= 1 - \(\frac{9}{10^{2013}+1}\)
Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\) nên 10.A > 10.B
=> A >B
Vậy ...........
\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
Vì \(1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\Rightarrow10A< 10B\)
\(\Rightarrow A< B\)
để so sánh A và B ta so sánh
\(\frac{10^{11}-1}{10^{12}-1}\)và \(\frac{10^{10}+1}{10^{11}+1}\)
Ta có \(10^{11}-1< 10^{11}+1\)
và \(10^{12}-1>10^{11}+1\)
=> A<B
\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
vì 1012-1>1011+1
=>\(\frac{9}{10^{12}-1}<\frac{9}{10^{11}+1}\)
=>A<B
Ta có:\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
Vì \(1-\frac{9}{10^{12}-1}<1+\frac{9}{10^{11}+1}\)
Nên A<B