Question

So sánh A=\(\frac{10^{11}-1}{10^{12}-1}\)và B = \(\frac{10^{10}+1}{10^{11}+1}\)

m.n ơi giúp mk với mk đang cần gấp

Answer 1

Ta có \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)

Vì \(\frac{9}{10^{12}-1}< \frac{9}{10^{11}+1};1=1\Rightarrow1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\Rightarrow\frac{10^{11}-1}{10^{12}-1}< \frac{10^{10}+1}{10^{11}+1}\)

Suy ra\(A< B\)

Answer 2

\(A=\frac{10^{11}-1}{10^{12}-1}\) => \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}\)

=> \(10A=1-\frac{9}{10^{12}-1}\)=> 10A < 1

\(B=\frac{10^{10}+1}{10^{11}+1}\) => \(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}\)

=> \(10B=1+\frac{9}{10^{11}+1}\)=> 10B > 1

=> 10B > 10A => B > A

ĐS: B > A

Answer 3

Ta có: 10A = \(\frac{10^{12}-10}{10^{12}-1}\)= 1 - \(\frac{9}{10^{12}-1}\) < 1 ( vì \(\frac{9}{10^{12}-1}\)\(\ne\)0 )
          10B = \(\frac{10^{11}+10}{10^{11}+1}\)= 1 + \(\frac{9}{10^{11}+1}\)> 1 ( vì \(\frac{9}{10^{11}+1}\)\(\ne\)0 )
Do đó: 10A < 1 < 10B
 \(\Rightarrow\)10A < 10B
 \(\Rightarrow\)A < B
Vậy: A < B.
- Đúng thì k cho mình nhé!!