A=\(7^{2018}\)- \(3^{2018}\)
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A = ( 157 x 2018 - 99 x 2018-2018 ) : 2018 - 57
= [2018 x ( 157 - 99 ) - 2018] : 2018 - 57
= [ 2018 x 58 - 2018 ] : 2018 - 57
= 115026 : 2018 - 57
= 57 - 57
= 0
B = ( 1 + 3 + 5 + 7 +...+ 2018 ) x ( 135135 x 137 - 135 x 137 )
( 1 + 3 + 5 + 7 +...+ 2018 ) = ( 2018 - 1 ) : 2 + 1 = 1009 số
( 1 + 3 + 5 + 7 +...+ 2018 ) = ( 2018 + 1 ) x 1009 : 2 = 1018585
Vậy B = ( 1 + 3 + 5 + 7 +...+ 2018 ) x ( 135135 x 137 - 135 x 137 )
B = 1018585 x 18495000 = số rất lớn
nếu bạn viết nhầm + thành x thì 1018585 + 18495000 = 19513588
nhớ k cho mình nhé !!!!
Ta có : A =\(\frac{2017}{2018}\)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 1
=\(\frac{2017}{2018}\)
Vậy A= : \(\frac{2017}{2018}\)
Bài giải
\(A=\frac{2017}{2018}\text{ x }\frac{7}{8}+\frac{2017}{2018}\text{ x }\frac{3}{8}-\frac{2017}{2018}\text{ x }\frac{1}{4}\)
\(A=\frac{2017}{2018}\text{ x }\frac{1}{4}\left(\frac{7}{2}+\frac{3}{2}-1\right)=\frac{2017}{2018}\text{ x }\frac{1}{4}\text{ x }4==\frac{2017}{2018}\text{ x }1=\frac{2017}{2018}\)
À mà thôi khỏi mình biết cách làm rồi ! Dù sao cũng cảm ơn lần nữa
(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
Tính à?
_____________________
2018S=2018+20182+20183+.....+20187+20188
2018S-S=20188-1
2017S=20188-1
=>S=..............tự làm tiếp.......................
a, (-45)+(-2018)+/-145/+/-2018/
=[(-45)+/-45/] +[(-2018)+/-2018/)
=(-45+45)+(-2018+2018)
=0
b, 5^3.4^3+85.5^3-5^7:5^4
=5^3.4^3+85.5^3-5^3
=5^3.(4^3+85-1)
=5^3.148
=18500
c, (7^108.50-7^108):7^110
=[7^108.(50-1)];7^110
=(49.7^108):7^110
=49:7^2
=49:49=1
a) (-45)+(-2018)+/-145/+/-2018/
=(-45)+(-2018)+145+2018
=(-45+145)+(-2018+2018)
=100+0
=100
b)53.42+85.53-57:54
=53.42+85.53-53
=53.(42+85-1)
=53.100
=12500
c) (7108.50-7108):7110
=7108.(50-1):7110
=7108.49:7110
=7108.72:7110
=7110:7110
=1
\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)
\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)
\(=2018\times0=0\)
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
Ta có:
72018-32018
=(74)504.72-(3504)4.32
=(...1).(...9)-(...1)-9
=(---9)-(..9)
=(..0)
Vì các số tận cùng là 0 thì chia hết cho 10 nên 72018-32018 chia hết cho 10 hay A chia hết cho 10
Vậy A chia hết cho 10
ta có:
7^2018-3^2018
theo công thức ta có:
a^b-c^b
=a-c^b
kết quả sẽ là:
7-3^2018
=4^2018
Ta có:A=72018 - 32018
=(7 - 3)2018
=42018
Vậy A=42018.