tính nhaanhcaau 2018x3+2018x8-2018
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2018 : 0,2 + 2018 x 0,5 + 2018 x 0,25 + 2018 x 89
= 2018 x 5 + 2018 x 0,5 + 2018 x 0,25 + 2018 x 89
= 2018 x ( 5 + 0,5 + 0,25 + 89 )
= 2018 x 94,75
= 191205,5
Study well ! >_<
2018 : 0,2 + 2018 x 0,5 + 2018 x 0,25 + 2018 x 89
= 2018 x 5 + 2018 x 0,5 + 2018 x 0,25 + 2018 x 89
= 2018 x ( 5 + 0,5 + 0,25 + 89 )
= 2018 x 94,75
= 191205,5
Study well ! >_<
2018 x 36 + 2018 + 2018 x 63
= 2018 x 36 + 2018 x 1 + 2018 x 63
= 2018 x ( 36 + 1 + 63 )
=2018 x 100
= 201800
^HT^
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
Tính bằng cách thuận tiện nhất:
2018 : 1/2 + 2018 :1/3 + 2018 : 1/4 +2018
= 2018 : 1/2 + 2018 :1/3 + 2018 : 1/4 +2018 : 1
= 2018 : (1/2 + 1/3 + 1/4 + 1 )
= 2018 : 25/12
= 968,64.
Mk ko bít đúng không nhưng mình cứ đăng lên, nếu sai thì thông cảm nhé.
2018x3+2018x8-2018x1
2018x(3+8-1)
2018x 10
20180
20180 nha <33