So sánh A=10^2016+2018/10^2017+2018 và B=10^2017+2018/10^2018+2018 Giải giúp mình với mai mình làm bài ktr rồi ah. Help 06/11/2021 Bởi Harper So sánh A=10^2016+2018/10^2017+2018 và B=10^2017+2018/10^2018+2018 Giải giúp mình với mai mình làm bài ktr rồi ah. Help
Đáp án:A=102016+2018102017+2018B=102017+2018102018+2018<1⇒102017+2018102018+2018<102017+2018+18162102018+2018+18262⇒B<102017+20180102018+20180⇒B<10(102016+2018)10.(102017+2018)⇒B<102016+2018102017+2018⇒B<A Giải thích các bước giải: Bình luận
Đáp án: $\begin{array}{l}A = \frac{{{{10}^{2016}} + 2018}}{{{{10}^{2017}} + 2018}}\\B = \frac{{{{10}^{2017}} + 2018}}{{{{10}^{2018}} + 2018}} < 1\\ \Rightarrow \frac{{{{10}^{2017}} + 2018}}{{{{10}^{2018}} + 2018}} < \frac{{{{10}^{2017}} + 2018 + 18162}}{{{{10}^{2018}} + 2018 + 18262}}\\ \Rightarrow B < \frac{{{{10}^{2017}} + 20180}}{{{{10}^{2018}} + 20180}}\\ \Rightarrow B < \frac{{10\left( {{{10}^{2016}} + 2018} \right)}}{{10.\left( {{{10}^{2017}} + 2018} \right)}}\\ \Rightarrow B < \frac{{{{10}^{2016}} + 2018}}{{{{10}^{2017}} + 2018}}\\ \Rightarrow B < A\end{array}$ Bình luận
Đáp án:A=102016+2018102017+2018B=102017+2018102018+2018<1⇒102017+2018102018+2018<102017+2018+18162102018+2018+18262⇒B<102017+20180102018+20180⇒B<10(102016+2018)10.(102017+2018)⇒B<102016+2018102017+2018⇒B<A
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
A = \frac{{{{10}^{2016}} + 2018}}{{{{10}^{2017}} + 2018}}\\
B = \frac{{{{10}^{2017}} + 2018}}{{{{10}^{2018}} + 2018}} < 1\\
\Rightarrow \frac{{{{10}^{2017}} + 2018}}{{{{10}^{2018}} + 2018}} < \frac{{{{10}^{2017}} + 2018 + 18162}}{{{{10}^{2018}} + 2018 + 18262}}\\
\Rightarrow B < \frac{{{{10}^{2017}} + 20180}}{{{{10}^{2018}} + 20180}}\\
\Rightarrow B < \frac{{10\left( {{{10}^{2016}} + 2018} \right)}}{{10.\left( {{{10}^{2017}} + 2018} \right)}}\\
\Rightarrow B < \frac{{{{10}^{2016}} + 2018}}{{{{10}^{2017}} + 2018}}\\
\Rightarrow B < A
\end{array}$