Ta luôn có nếu a>0; b>0 thì \(\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Áp dụng vào bài toán ta thấy 1011-1 > 0 và 1012-1 > 0 nên
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
Vậy A < B
Xin lỗi bn nhé bài toán phụ phía trên đang còn 1 đk nữa là a<b
\(10A=\frac{10^{12}-10}{10^{12}-1}\); \(10B=\frac{10^{11}+10}{10^{11}+1}\)
=> \(10A=\frac{\left(10^{12}-1\right)-9}{10^{12}-1}\); \(10B=\frac{\left(10^{11}+1\right)+9}{10^{11}+1}\)
=> \(10A=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}\); \(10B=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}\)
=> \(10A=1-\frac{9}{10^{12}-1}\); \(10B=1+\frac{9}{10^{11}+1}\)
Ta có: \(10A=1-\frac{9}{10^{12}-1}< 1\); \(10B=1+\frac{9}{10^{11}+1}>1\)
=> \(10A< 10B\)
=> \(A< B\)
\(A-B=\frac{10^{11}-1}{10^{12}-1}-\frac{10^{10}+1}{10^{11}+1}\)
\(\Leftrightarrow\frac{\left(10^{11}-1\right)\left(10^{11}+1\right)-\left(10^{10}+1\right)\left(10^{12}-1\right)}{\left(10^{12}-1\right)\left(10^{11}+1\right)}\)
\(\Leftrightarrow\frac{10^{22}-1-10^{22}+10^{10}-10^{12}+1}{\left(10^{12}-1\right)\left(10^{11}+1\right)}\)
\(\Leftrightarrow\frac{10^{10}-10^{12}}{\left(10^{12}-1\right)\left(10^{11}+1\right)}< 0\)
\(\Leftrightarrow A-B< 0\)
\(\Leftrightarrow A< B\)
:))
ta thấy: A=\(\frac{10^{11}-1}{10^{12}-1}< 1\)
Ta có: nếu\(\frac{a}{b}< 1\)thì \(\frac{a+n}{b+n}>\frac{a}{b}\Rightarrow A< \frac{\left(10^{11}-1\right)+11}{\left(10^{12}-1\right)+11}\)=\(\frac{10^{11}+10}{10^{12}+10}\)
Do đó A<\(\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\)=\(\frac{10^{10}+1}{10^{11}+1}\)
Vậy A<B