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Ta có: \(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)\(\Rightarrow10A=\frac{10^{2017}+2018.10}{10^{2017}+2018}=\frac{10^{2017}+2018+2018.9}{10^{2017}+2018}=1+\frac{2018.9}{10^{2017}+2018}\)
Tương tự ta có: \(10B=1+\frac{2018.9}{10^{2018}+2018}\)
Vì \(2017< 2018\)\(\Rightarrow10^{2017}< 10^{2018}\)\(\Rightarrow10^{2017}+2018< 10^{2018}+2018\)
\(\Rightarrow\frac{2018.9}{10^{2017}+2018}>\frac{2018.9}{10^{2018}+2018}\)\(\Rightarrow1+\frac{2018.9}{10^{2017}+2018}>1+\frac{2018.9}{10^{2018}+2018}\)
hay \(10A>10B\)\(\Rightarrow A>B\)
Vậy \(A>B\)
Ta có : \(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}=\frac{10^{2017}+2018+18162}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\)
Ta có : \(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)
\(\Rightarrow\frac{10^{2018}+20180}{10^{2018}+2018}=\frac{10^{2018}+2018+18162}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\)
Vì \(10^{2017}+2018< 10^{2018}+2018\) nên \(\frac{18162}{10^{2017}+2018}>\frac{18162}{10^{2018}+2018}\)
\(\Rightarrow1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2017}+2018}\Rightarrow10A>10B\Rightarrow A>B\)
Vậy A > B
Làm khác bạn kia 1 xíu à
\(+)A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(10A=\frac{10^{2017}+20180}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\left(1\right)\)
\(+)10B=\frac{10^{2018}+20180}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\left(2\right)\)
Từ (1),(2)=> \(\frac{18162}{10^{2017}+2018} >\frac{18162}{10^{2018}+2018}\)
=> 10A>10B
=>A>B
\(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}\)
\(=\frac{10^{2017}+2018+18162}{10^{2017}+2018}\)
\(=\frac{10^{2017}+2018}{10^{2017}+2018}+\frac{18162}{10^{2017}+2018}\)
\(=1+\frac{18162}{10^{2017}+2018}\)
\(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)
\(\Rightarrow10B=\frac{10^{2018}+20180}{10^{2018}+2018}\)
\(=\frac{10^{2018}+2018+18162}{10^{2018}+2018}\)
\(=\frac{10^{2018}+2018}{10^{2018}+2018}+\frac{18162}{10^{2018}+2018}\)
\(=1+\frac{18162}{10^{2018}+2018}\)
Ta thấy: \(1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2018}+2018}\)
=> 10A > 10B
=> A > B
dễ mà bạn
A=10x10+10/ 10x10x10+10
A=110/1010
a=11/101
b=10x10-10/10x10x10-10
b=90/990
b=11/110
vậy a=11/101
b=90/990
bn tự so sánh nhé ^-^
mik mỏi tay quá ko đánh đc nữa bọn mik bằng tuổi đó
câu này mik học trên lớp rùi
2017^10+2017^9=2017^9+(1+2017)=2017^9x2018
2018^10=2018^9+2018
ta có:2018=2018
2017^9<2018^9
=>2017^10+2017^9<2018^10
Ta có : \(A=\frac{10^{2016}+1}{10^{2017}+1}\)
Suy ra \(10A=\frac{10^{2017}+10}{10^{2017}+1}\)
Suy ra \(10A=1+\frac{9}{10^{2017}+1}\)
Ta lại có : \(B=\frac{10^{2017}+1}{10^{2018}+1}\)
Suy ra : \(10B=\frac{10^{2018}+10}{10^{2018}+1}\)
Suy ra : \(10B=1+\frac{9}{10^{2018}+1}\)
Vì \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\)
Nên \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
Suy ra \(10A>10B\)
Suy ra \(A>B\)
nhanh lên các bn mik cần gấp