tính bằng cách thuận tiện nhất : 2018+2018+2018+2018+2018*7-2018
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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^
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.
(Dấu . là dấu nhân)
a/\(\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}:3\)
\(=\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\cdot1\)
\(=\dfrac{2}{5}\)
b/\(\dfrac{2010}{2018}:\dfrac{1}{2}+\dfrac{7}{2018}:\dfrac{1}{2}\)
\(=\left(\dfrac{2010}{2018}+\dfrac{7}{2018}\right):\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}:\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}\cdot2\)
\(=\dfrac{2017}{1009}\)
a, \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) : 3
= \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) \(\times\) \(\dfrac{1}{3}\)
= \(\dfrac{2}{5}\) \(\times\) ( \(\dfrac{4}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{2}{5}\) \(\times\) 1
= \(\dfrac{2}{5}\)
b, \(\dfrac{2010}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{7}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{1}{2018}\) : \(\dfrac{1}{2}\)
= \(\dfrac{2010}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{7}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{1}{2018}\) \(\times\) \(\dfrac{2}{1}\)
= \(\dfrac{2}{1}\) \(\times\) ( \(\dfrac{2010}{2018}\) + \(\dfrac{7}{2018}\) + \(\dfrac{1}{2018}\))
= 2 \(\times\) \(\dfrac{2018}{2018}\)
= 2 \(\times\) 1
= 2
\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)
\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)
\(=2018\times0=0\)
B-a:tương tự bài A
b,\(\frac{14}{18}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)
\(=\frac{7}{9}\times1=\frac{7}{9}\)\
thanks
B-a,\(\frac{3\times125+3\times125}{6\times43+6\times57}=\frac{2\times3\times125}{6\times\left(43+57\right)}\)
\(=\frac{3\times250}{6\times100}=\frac{5}{2\times2}=\frac{5}{4}\)
thanks
\(\frac{2017}{2018}\)x\(\frac{7}{8}\)+\(\frac{2017}{2018}\)x\(\frac{3}{8}\)-\(\frac{2017}{2018}\)x\(\frac{1}{4}\)
= \(\frac{2017}{2018}\)x (\(\frac{7}{8}\)+\(\frac{3}{8}\)-\(\frac{1}{4}\))
= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{1}{4}\))
= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{2}{8}\))
= \(\frac{2017}{2018}\)x 1
= \(\frac{2017}{2018}\)
Chúc em học tốt nhé :>
=2017/2018*(7/8+3/8)-2017*1/4
=2017/2018*5/4+2017*-1/4
=2017/2018*(5/4-1/4)
=2017/2018*1
=2017/2018
\(2018\cdot2018-2017\cdot2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)\)
\(=2018^2-2018^2+1\)
\(=1\)
\(2018.2018-2017.2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)\)
\(=2018^2-2018^2+1\)
\(=1\)
Bài 1:a, Tính bằng cách thuận tiện nhất :
2018 x 32 + 8072 : 4 x 23 + 4036 : 2 x 25 + 2018 +2018 x 19
=2018 x 32 + 2018 x 23 + 2018 x 25 + 2018 x 1 + 2018 x 19
= 2018 x ( 32 + 23 + 25 + 1 + 19 )
= 2018 x 100
=201800
Bài 2 :a, Tính bằng cách thuật tiện :
3 x 4/15 + 2 x 4/15 - 5 x 4/15
= ( 3 x 2 - 5 ) x 4/15
= 0 x 4/15
= 0
2018 + 2018 + 2018 + 2018 + 2018*7 – 2018
= 2018 x 4 + 2018 x 7 – 2018
= 2018 x (4 + 7) – 2018
= 2018 x 11 – 2018 = 2018 x 10 = 20180
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