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24 tháng 3 2020

50%=1/2

0,25=1/4

50% *x + 3/4 * x + 0,25 * x - 1/2 * x = 2018

=> (1/2 + 3/4 + 1/4 - 1/2)* x = 2018

=> 1 * x = 2018

=> x = 2018

24 tháng 3 2020

\(50\%\cdot x+\frac{3}{4}\cdot x+0,25-x+\frac{1}{2}\cdot x=2018\)

\(\Leftrightarrow0,5x+\frac{3}{4}x+0,25x+\frac{1}{2}x=2018\)

\(\Leftrightarrow x\left(0,5+\frac{3}{4}+0,25+\frac{1}{2}\right)=2018\)

\(\Leftrightarrow1,75x=2018\)

\(\Leftrightarrow x=...\)
HỌC TỐT!
Cơ mà "Toán lớp 5"????

22 tháng 4 2019

Sửa đề: \(\frac{x-3}{2018}\rightarrow\frac{x-3}{2016}\)

\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)

\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)

\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-2019=0\) (Vì \(\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)\ne0\) )

\(\Leftrightarrow x=2019\)

Vậy \(S=\left\{2019\right\}\)

18 tháng 8 2019

Này Thục Trinh, chỗ mà \(\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)

bạn phải có đóng mở ngoặc vào chứ

8 tháng 7 2017

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

1 tháng 8 2017

khó lắm

bây h thì bạn giải đc chưa

7 tháng 8 2018

help me

7 tháng 8 2018

\(a)\) Ta có : 

\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

22 tháng 7 2018

\(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)

\(\Rightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)

\(\Rightarrow\frac{x+4+2016}{2016}+\frac{x+3+2017}{2017}=\frac{x+2+2018}{2018}+\frac{x+1+2019}{2019}\)

\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{2018}+\frac{x+2020}{2019}\)

\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)

\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)

\(\Rightarrow x+2020=0\) vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)

\(\Rightarrow x=-2020\)

17 tháng 8 2019

\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)

\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)

\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}=\frac{x-2020}{2017}+\frac{x-2020}{2016}\)

\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)

\(\Leftrightarrow\left(x-2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

\(\Leftrightarrow x-2020=0\)

\(\Leftrightarrow x=0+2020\)

\(\Rightarrow x=2020\)

Vậy \(x=2020.\)

Chúc bạn học tốt!

16 tháng 7 2018

<=>[ (x-1)/2019] -1 +[(x-2)/2018]-1 = [(x-3)/2017]-1 +[(x-4)/2016] -1

<=> (x-2020)/2019 +(x-2020)/2018 = (x-2020)/2017 + (x-2020)/2016

<=> (x-2020)( 1/2019+1/2018-1/2017-1/2016)= 0

=> x-2020= 0 => x= 2020

\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)

\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)

\(\Rightarrow\frac{x-1-2018}{2018}+\frac{x-2-2017}{2017}=\frac{x-3-2016}{2016}+\frac{x-4-2015}{2015}\)

\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)

\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)

\(\Rightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

Mà \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\)

\(\Rightarrow x-2019=0\)

\(\Rightarrow x=2019\)