So sánh A = 10^99 +1/ 10^98+1 B= 10^98+1/10^97+1 18/08/2021 Bởi Mackenzie So sánh A = 10^99 +1/ 10^98+1 B= 10^98+1/10^97+1
Đáp án: Ta có : A = $\frac{10^{99}+1}{10^{98}+1}$ => $\frac{A}{10}$ = $\frac{10^{99}+1}{10^{99}+10}$ = 1 – $\frac{9}{10^{99}+10}$ B = $\frac{10^{98}+1}{10^{97}+1}$ => $\frac{B}{10}$ = $\frac{10^{98}+1}{10^{98}+10}$ = 1 – $\frac{9}{10^{98}+10}$ Do $10^{99} > 10^{98}$ $=> 10^{99} + 10 > 10^{98} + 10$ $=> \frac{9}{10^{99}+10} < \frac{9}{10^{98}+10}$ => $1 – \frac{9}{10^{99}+10} > 1- \frac{9}{10^{98}+10}$ => $\frac{A}{10}$ > $\frac{B}{10}$ Giải thích các bước giải: Bình luận
`\text{Ta có} `: `A > 1` `\text{( tử > mẫu )}` `A = \frac{10^{99} + 1}{10^{98} + 1} > \frac{10^{99} + 1 + 9}{10^{98} + 1 + 9} = \frac{10^{99} + 10}{10^{98} +10} = \frac{10(10^{98} + 1)}{10(10^{97} + 1 )} = \frac{10^{98} + 1}{10^{97} + 1} = B` `=> A > B` Bình luận
Đáp án:
Ta có :
A = $\frac{10^{99}+1}{10^{98}+1}$
=> $\frac{A}{10}$ = $\frac{10^{99}+1}{10^{99}+10}$ = 1 – $\frac{9}{10^{99}+10}$
B = $\frac{10^{98}+1}{10^{97}+1}$
=> $\frac{B}{10}$ = $\frac{10^{98}+1}{10^{98}+10}$ = 1 – $\frac{9}{10^{98}+10}$
Do $10^{99} > 10^{98}$
$=> 10^{99} + 10 > 10^{98} + 10$
$=> \frac{9}{10^{99}+10} < \frac{9}{10^{98}+10}$
=> $1 – \frac{9}{10^{99}+10} > 1- \frac{9}{10^{98}+10}$
=> $\frac{A}{10}$ > $\frac{B}{10}$
Giải thích các bước giải:
`\text{Ta có} `: `A > 1` `\text{( tử > mẫu )}`
`A = \frac{10^{99} + 1}{10^{98} + 1} > \frac{10^{99} + 1 + 9}{10^{98} + 1 + 9} = \frac{10^{99} + 10}{10^{98} +10} = \frac{10(10^{98} + 1)}{10(10^{97} + 1 )} = \frac{10^{98} + 1}{10^{97} + 1} = B`
`=> A > B`