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

=>x+2019=0

=>x=-2019

13 tháng 12 2018

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

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

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

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

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

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

\(\Leftrightarrow x+2019=0\)

\(\Leftrightarrow x=-2019\)

Vậy...

18 tháng 10 2017

\(\dfrac{x-2}{2018}=\dfrac{x-3}{2017}=\dfrac{x-4}{2016}=\dfrac{x-5}{2015}\)

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

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

\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=\dfrac{x-2020}{2016}+\dfrac{x-2020}{2015}\)

\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}-\dfrac{x-2020}{2016}-\dfrac{x-2020}{2015}=0\)

\(\left(x-2020\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)

\(\dfrac{1}{2018};\dfrac{1}{2017};\dfrac{1}{2016};\dfrac{1}{2015}>0\)

Nên \(x-2020=0\)

\(x=0+2020\)

\(x=2020\)

Vậy x bằng 2020

21 tháng 10 2017

Tui đánh giá cao câu trả lời này của bạn :v

5 tháng 3 2023

\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)

\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)

 

24 tháng 6 2018

Giải:

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

\(\Leftrightarrow2+\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=2+\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)

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

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

\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}=\dfrac{x+2020}{3}+\dfrac{x+2020}{2}\)

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

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

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy ...

\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)

\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)

\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)

\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)

23 tháng 9 2021

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

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

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

\(\Leftrightarrow x=-2020\)(do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\ne0\))

23 tháng 9 2021

Cộng 1 vào mỗi số hạng là ra

21 tháng 4 2018

\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)

\(\Rightarrow\dfrac{x-1}{2017}+\dfrac{x-2}{2016}-\dfrac{x-3}{2015}-\dfrac{x-4}{2014}=0\)

\(\Rightarrow\dfrac{x-1}{2017}-1+\dfrac{x-2}{2016}-1-\dfrac{x-3}{2015}+1-\dfrac{x-4}{2014}+1=0\)

\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)-\left(\dfrac{x-3}{2015}-1\right)-\left(\dfrac{x-4}{2014}-1\right)=0\)

\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)

\(\Rightarrow x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

\(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)

Để \(x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

\(\Rightarrow x-2018=0\)

\(x=2018\)

21 tháng 4 2018

Ta có :

\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)

\(\Leftrightarrow\)\(\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\) ( trừ 2 vế cho 2 )

\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)

\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)

\(\Leftrightarrow\)\(\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

\(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)

Nên \(x-2018=0\)

\(\Rightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~

a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)

\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)

=>x=13/12

b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)

\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)

\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)

c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)

\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)

\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)

d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)

\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)

\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)

\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)

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

=>x+2020=0

hay x=-2020

a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

=>16x-7=13x+2

=>3x=9

hay x=3

b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)

=>x+2017=0

hay x=-2017

e: \(\left(2x-3\right)^2=144\)

=>2x-3=12 hoặc 2x-3=-12

=>2x=15 hoặc 2x=-9

=>x=15/2 hoặc x=-9/2