Bài 2: Tính bằng cách thuận tiện:
a, 9012 x 999 + 9012 B, 2018 x 2017 + 2018 - 2018 x 2008
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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
\(\frac{2017}{2018}\)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
\(2018\cdot2018-2017\cdot2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)\)
\(=2018^2-2018^2+1\)
\(=1\)
\(2018.2018-2017.2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)\)
\(=2018^2-2018^2+1\)
\(=1\)
(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
a) 4/15 x 2016/2018+ 4/15 x 3/2018- 4/15 x 1/2018
= 4/15 x (2016/2018 + 3/2018 - 1/2018)
= 4/15 x (\(1\frac{1}{2018}\)-1/2018)
=4/15 x 1
= 4/15
b) 8/15 x 8 - 2 x 8/15+3 x 8/15+ 8/15
= 8/15 x ( 8 - 2 + 3 + 1)
= 8/15 x 10
= 80/15 => \(5\frac{1}{3}\)
Mấy cái kia mình lười làm phân số nên mình chỉ viết rõ hỗn số thôi, bạn thông cảm nha. Chúc bạn học giỏi <3
9012 x 999 + 9012
= 9012 x 999 + 9012 x 1
= 9012 x ( 999 + 1)
= 9012 x 1000
= 9012000
2018 x 2017 + 2018 - 2018 x 2008
= 2018 x 2017 + 2018 x 1 - 2018 x 2008
= 2018 x ( 2017 + 1 - 2018)
= 2018 x ( 2018 - 2018)
= 2018 x 0
= 0
a, 9012 x 999 + 9012
= 9012 x (999+1)
= 9012 x 1000
= 9012000
b,
2018 x 2017 + 2018 - 2018 x 2008
= 2018 x (2017 + 1 - 2008)
= 2018 x 10
= 20 180