(-2018)/2019.2/7-2018/2019.5/7:1_2018//2019
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Cho a<b ,chứng tỏ rằng 2020 -2019a>2018-2019b
Mk làm theo cách này,các bn xm đc k nhes
Vì a<b(gt) ,thay a=1,b=2 vào bđt 2020-2019 và 2018-2019b ta có :
VT=2020-2019a=2010-2019.1
=>VT=1
VP=2018-2019b=2018-2019.2
=>VP=-2020
Vì 1>-2010 nên VT>VP
Vậy 2020 -2019a>2018-2019b(dpcm)
Ko thể dùng 1 trường hợp cụ thể để chứng minh dạng tổng quát.
Cách chứng minh bài này rất đơn giản:
\(a< b\Rightarrow2019a< 2019b\)
\(\Rightarrow-2019a>-2019b\)
\(\Rightarrow-2019a+2020>-2019b+2020>-2019b+2018\)
Vậy \(2020-2019a>2018-2019b\)
cho S=1 + 2018 + 20182 +20183 + 20184 + 20185+ 20186 + 20187
Tính à?
_____________________
2018S=2018+20182+20183+.....+20187+20188
2018S-S=20188-1
2017S=20188-1
=>S=..............tự làm tiếp.......................
5+5/23.5/27+5/2018
________________
7+7/23.7/27+7/2018
=5.(1+1/23+1/27+1/2018)
_______________________=5/7
7.(1+1/23+1/27+1/2018)
\(\frac{5.\left(1+\frac{1}{23}+\frac{1}{27}+\frac{1}{2018}\right)}{7.\left(1+\frac{1}{23}+\frac{1}{27}+\frac{1}{2018}\right)}\)
=5/7
a=\(\frac{-7}{2018^{2019}}\)+ \(\frac{-15}{2018^{2018}}\)
b= \(\frac{-15}{2018^{2019}}\)+ \(\frac{-7}{2018^{2018}}\)
mk cần gấp nha
a=\(\frac{-7}{2018^{2019}}+\frac{-15}{2018^{2018}}=\frac{-7-15.2018}{2018^{2019}}\)
b=\(\frac{-15}{2018^{2019}}+\frac{-7}{2018^{2018}}=\frac{-15-7.2018}{2018^{2019}}\)
Ta có: -7-15.2018 < -15-7.2018
\(\Rightarrow\frac{-7-15.2018}{2018^{2019}}< \frac{-15-7.2018}{2018^{2019}}\)
\(\Rightarrow a< b\)
a)A=/x+7/+/x^2-169/-/x-2018/
b)B=[2018/2+2018/3+2028/4+.....+2019/2018]:[1/2018+2/2017+3/2016+......+2018]
Thực hiện phép tính:\(\frac{-2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\frac{2018}{2019}\)
\(-\frac{2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\frac{2018}{2019}=\frac{2018}{2019}\left(\frac{-2-5}{7}\right)+1\frac{2018}{2019}=\frac{2018}{2019}.\left(-1\right)+1\frac{2018}{2019}=\frac{-2018}{2019}+1\frac{2018}{2019}=1\)
Tinh:A=2010^2010.(7^10:7^8-3.2^4-2^2018:2^2018)
A = 2017/2018 x 7/8 + 2017/2018 x 3/8 - 2017/2018 x 1/4
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
\(A=\frac{2017}{2018}\text{ x }\frac{7}{8}+\frac{2017}{2018}\text{ x }\frac{3}{8}-\frac{2017}{2018}\text{ x }\frac{1}{4}\)
\(A=\frac{2017}{2018}\text{ x }\frac{1}{4}\left(\frac{7}{2}+\frac{3}{2}-1\right)=\frac{2017}{2018}\text{ x }\frac{1}{4}\text{ x }4==\frac{2017}{2018}\text{ x }1=\frac{2017}{2018}\)
Gấp lắm bạn !!!?!!!!
a. A=(157×2018-99×2018-2018) : 2018-57
b. B=(1+3+5+7+...+2018) × (135135×137-135×137
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 !!!!
a)= (2018 . (157-99-1)):2018-57
=2018.57:2018-57
=????
Tính bằng cách thuận tiện:
A = 2017/2018 x 7/8 + 2017/2018 x 3/8 - 2017/2018 x 1/4