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\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)
\(\Leftrightarrow4037(x+2019)=4038(x+2018)\)
\(\Leftrightarrow4037x+8150703=4038x+8148684\)
\(\Leftrightarrow4037x+8150703-4038x=8148684\)
\(\Leftrightarrow4037x-4038x+8150703=8148684\)
\(\Leftrightarrow-x=-2019\)
\(\Leftrightarrow x=2019\)
(x+2019) x 4037=(x+2018) x 4038
⇒4037�+(4037×2019)=4038�+(4038×2018)⇒4037x+(4037×2019)=4038x+(4038×2018)
⇒4037�+8150703=4038�+8148684⇒4037x+8150703=4038x+8148684
⇒4037�−4038�=−8150703+8148684⇒4037x−4038x=−8150703+8148684
⇒−�=−2019
⇒�=2019⇒x=2019
x+2019) x 4037=(x+2018) x 4038
⇒4037�+(4037×2019)=4038�+(4038×2018)⇒4037x+(4037×2019)=4038x+(4038×2018)
⇒4037�+8150703=4038�+8148684⇒4037x+8150703=4038x+8148684
⇒4037�−4038�=−8150703+8148684⇒4037x−4038x=−8150703+8148684
⇒−�=−2019
⇒�=2019⇒x=2019
1 tick với nha
\(\frac{x+2019}{x+2018}=\frac{4038}{4037}\)
\(\Rightarrow\left(x+2019\right)4037=\left(x+2018\right)4038\)
\(\Rightarrow4037x+\left(4037\times2019\right)=4038x+\left(4038\times2018\right)\)
\(\Rightarrow4037x+8150703=4038x+8148684\)
\(\Rightarrow4037x-4038x=-8150703+8148684\)
\(\Rightarrow-x=-2019\)
\(\Rightarrow x=2019\)
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.
1/3 + 1/6 + x/2018 = 1
<=> 1/2 + x/2018 = 1
<=> x/2018 = 1 - 1/2
<=> x/2018 = 1/2
<=> x = 2018 x 1/2
<=> x = 1009
Không hiểu thì ib tớ giải thích
\(\frac{1}{3}+\frac{1}{6}+\frac{x}{2018}=1\)
\(\Leftrightarrow\frac{1}{6}+\frac{x}{2018}=1-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}+\frac{x}{2018}=\frac{3}{3}-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}+\frac{x}{2018}=\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{2018}=\frac{2}{3}-\frac{1}{6}\)
\(\Leftrightarrow\frac{x}{2018}=\frac{4}{6}-\frac{1}{6}\)
\(\Leftrightarrow\frac{x}{2018}=\frac{3}{6}\)
\(\Leftrightarrow\frac{x}{2018}=\frac{1}{2}\)
\(\Leftrightarrow\frac{x}{2018}=\frac{1009}{2018}\)
\(\Leftrightarrow x=1009\)
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
\(\dfrac{x+2019}{x+2018}=\dfrac{4038}{4037}\)
=>\(\dfrac{x+2018+1}{x+2018}=1+\dfrac{1}{4037}\)
=>\(1+\dfrac{1}{x+2018}=1+\dfrac{1}{4037}\)
=>x+2018=4037
=>x=2019
x+2019=40374038
=>𝑥+2018+1𝑥+2018=1+14037x+2018x+2018+1=1+40371
=>1+1𝑥+2018=1+140371+x+20181=1+40371
=>x+2018=4037
=>x=2019