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\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)
\(\Rightarrow\left(x+2019\right)4037=\left(x+2018\right)4038\)
\(\Rightarrow4037x+\left(4037\times2019\right)=4038x+\left(4038\times2018\right)\)
\(\Rightarrow4037x+8150703=4038x+8148684\)
\(\Rightarrow4037x-4038x=-8150703+8148684\)
\(\Rightarrow-x=-2019\)
\(\Rightarrow x=2019\)
P/s: Số to kinh -_- Ko chắc đúng đâu.
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
\(\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019.2020-2.2019}{2019\left(2017+1\right)}=\frac{2019\left(2020-2\right)}{2019.2018}=\frac{2019.2018}{2019.2018}=1\)
\(A=\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019\left(2020-2\right)}{2019\left(2017+1\right)}\)
\(=\frac{2019.2018}{2019.2018}=1\)
Vậy \(A=1.\)
Mà lớp 5 làm gì đã học đến dấu \(.\)(dấu nhân lớp 5 viết kiểu này cơ: x )
Chúc em học tốt.
a) \(\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{2}:\frac{3}{2}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2017\times2018+2018+2019\right)\times0\)
\(=0\)
b) 10,11 + 11,12 + 12,13 + ...+ 98,99 + 99, 100
Số số hạng từ 10,11 đến 98,99 là:
( 98,99 - 10,11) : 1,01 + 1= 89
Tổng dãy số trên từ 10,11 đến 98,99 là:
( 98,99 + 10,11) x 89 : 2 = 4 854,95
=> 10,11 + 11,12+12,13 + ...+ 98,99+ 99,100 = 4 854,95 + 99, 1 = 4 954, 05
tìm x
( X x 0,25 + 2018 ) x 2019 = ( 51 + 2018 ) x 2019
( X x 0,25 + 2018 ) x 2019 = 2069 x 2019
( X x 0,25 + 2018 ) x 2019 = 4177311
X x 0,25 + 2018 = 4177311 : 2019
X x 0,25 + 2018 = 2069
X x 0,25 =2069 - 2018
X x 0,25 = 51
X = 51 : 0,25
X = 204
học tốt ^-^
Ta có : \(\frac{1}{n}+\frac{2020}{2019}=\frac{2019}{2018}+\frac{1}{n+1}\)
=> \(\frac{1}{n}-\frac{1}{n+1}=\frac{2019}{2018}-\frac{2020}{2019}\)
=> \(\frac{n+1}{n\left(n+1\right)}-\frac{n}{\left(n+1\right)n}=\frac{1}{4074342}\)
=> \(\frac{1}{n\left(n+1\right)}=\frac{1}{2018.2019}\)
=> n(n + 1) = 2018.2019
=> n(n + 1) = 2018.(2018 + 1)
=> n = 2018
x = 4038 - 2019 = 2019
~Học tốt~
#ngocyen#
\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)
\(\Leftrightarrow4037(x+2019)=4038(x+2018)\)
\(\Leftrightarrow4037x+8150703=4038x+8148684\)
\(\Leftrightarrow4037x+8150703-4038x=8148684\)
\(\Leftrightarrow4037x-4038x+8150703=8148684\)
\(\Leftrightarrow-x=-2019\)
\(\Leftrightarrow x=2019\)