Gấp lắm bạn !!!?!!!!
a. A=(157×2018-99×2018-2018) : 2018-57
b. B=(1+3+5+7+...+2018) × (135135×137-135×137
Tinh nhanh :
A = ( 2016 + 2017 ) - ( 2017 + 2018 ) + ( 2018 - 16 )
B = ( 157 - 215 ) + ( 315 - 157 ) + ( 215 - 265 )
C = 1 . 3 - 2 . 4 + 3 . 5 + 4 . 6 + ..... + 7 . 9.- 8 . 10
A = ( 2016 + 2017 ) - ( 2017 + 2018 ) + ( 2018 - 16 )
A = 2016 + 2017 - 2017 - 2018 + 2018 - 16
A = ( 2016 - 16 ) + ( 2017 - 2017 ) + ( 2018 - 2018 )
A = 2000 + 0 + 0
A = 2000
B = ( 157 - 215 ) + ( 315 - 157 ) + ( 215 - 265 )
B = 157 - 215 + 315 - 157 + 215 - 265
B = ( 157 - 157 ) + ( 215 - 215 ) + ( 315 - 265 )
B = 0 + 0 + 50
B = 50
cam on ban minh rat can cau C giup minh nhe !
Bài 3: Không quy đồng hãy so sánh các phân số sau: a, 2019/2020 và 2021/2022 b, 2019/2017 và 2021/2019 c, 201/202 và 135/137 d, 2019/2018 và 2021/2019
GẤP ... GẤP ... GẤP CÁC BẠN
P = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{4003}{\left(2016.2017\right)^3}\)
Chứng minh rằng : P < 1
A = \(\frac{2018^{100}+2018^{96}+...+2018^4+1}{2018^{102}+2018^{100}+...+2018^2+1}\)
Chứng minh rằng : 4A < \(10111^6\)
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]
Tính bằng cách thuận tiện nhất a) 2/5x4/3-2/5:3 b) 2010/2018:1/2+7/2018:1/2+1/2018:1/2
(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=2017^2018+1/2017^2018-3
B=2017^2018-1/2017^2018-5
So sánh
A = 2018 99 -1 phần 2018100 - 1
B = 201898 -1 phần 201899 - 1
33...3 (2018 số 3) . 33...3 (2018 số 3)
= 3 . 11...1 (2018 số 1) . 33...3 (2018 số 3)
= 11...1 (2018 số 1) . 99...9 (2018 số 9)
= 11...1 (2018 số 1) . (100...0 (2018 số 0) - 1)
= 11...1 (2018 số 1) . 100...0 (2018 số 0) - 11...1 (2018 số 1)
TIẾP THEO LÀ GÌ THÌ CÁC BẠN GIÚP MÌNH NHÉ VÌ PHẦN SAU MÌNH KO BIẾT LÀM. MÌNH CẢM ƠN NHIỀU!!!!
ê\(rrwwr\hept{\begin{cases}\\\end{cases}rrw^2r}\)
a)so sanh E=2018^99-1/2018^100-1 va F=2018^98/2018^99-1
b)Tim so nguyen to ab (a>b>0),biet ab-ba la so chinh phuong
c)Cho abc la so tu nhien co 3 chu so .Tim gia tri lon nhat cua A={abc/a+b+c}+1918