So sanh A và B :
\(A=\frac{10^{15}+1}{10^{16}+1}\); \(B=\frac{10^{16}+1}{10^{17}+1}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trước hết ta so sánh 10A và 10B
Ta có:
\(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\) \(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì: \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) nên 10A > 10B, do đó A>B
Ta thấy:B<1 vì 1015+1<1016+1
Theo quy tắc :\(\frac{a}{b}\)<\(\frac{a+m}{b+m}\)nên ta có: B =\(\frac{10^{16}+1}{10^{17}+1}\)<\(\frac{10^{16}+1+9}{10^{17}+1+9}\)<\(\frac{10^{16}+10}{10^{17}+10}\)<\(\frac{10\left(10^{15}+1\right)}{10\left(10^{16}+1\right)}\)=A
Suy ra B<A
![](https://rs.olm.vn/images/avt/0.png?1311)
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Nhận thấy: \(\frac{9}{10^{17}+1}< \frac{9}{10^{16}+1}\)=> 10B < 10A
=> A > B
A = ( 10^15+1 ) / ( 10^16+1 ) => 10A = ( 10^16+10 ) / ( 10^16+1 ) = 1 + ( 9/10^15+1 )
B = ( 10^16+1 ) / ( 10^17+1 ) => 10B = ( 10^17+10 ) / ( 10^17+1 ) = 1 + ( 9/10^16+1 )
Vì 10^15+1 < 10^16+1 nên 9/10^15+1 > 9/10^16+1 => 1 + ( 9/10^15+1 ) > 1 + ( 9/10^16+1 )
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(10^{16}+1< 10^{17}+1\) nên \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) \(\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
=> 10A > 10B Do đó A > B
Vậy A > B
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
Ta có:
\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{\left(10^{15}+1\right).10}{\left(10^{16}+1\right).10}=\frac{10^{16}+10}{10^{17}+10}=\frac{10^{16}+1+9}{10^{17}+1+9}\)
Vì \(B=\frac{10^{16}+1}{10^{17}+1}< 1\)
\(\Rightarrow B=\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}=A\)
Vậy B < A
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(\Rightarrow10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(A=\frac{10^{16}+1}{10^{17}+1}\)
\(\Rightarrow10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\left(Do10^{16}+1< 10^{17}+1\right)\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
sai đề rồi bạn.\(\frac{a}{b}>\frac{a+c}{b+c}\) với \(a>b\) mới đúng nha.
Ta có:\(A=\frac{10^{17}+1}{10^{16}+1}>\frac{10^{17}+1+9}{10^{16}+1+9}=\frac{10^{17}+10}{10^{16}+10}=\frac{10\left(10^{16}+1\right)}{10\left(10^{15}+1\right)}=\frac{10^{16}+1}{10^{15}+1}\)
\(\Rightarrow A>B\)
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=\frac{1+9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=\frac{1+9}{10^{17}+1}\)
Vì\(\frac{1+9}{10^{16}+1}\)>\(\frac{1+9}{10^{17}+1}\)
=>10A>10B
=>A>B