\(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{3x+12}{2015}+3\)

\(\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{3 \left(x+2019\right)}{2015}\)

\(\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\right)=0\)

mà \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\ne0\Rightarrow x+2019=0\Leftrightarrow x=-2019\)

Ai biết làm giải giùm mình với

Bạn có chắc chắn là đúng không

Đáp án là : 

Tìm x : 

x = -2019 

x= -2019

Nguyễn Tiến Đạt

a)\(|3x-5|=|x+2|\)

=> Ta có 2 trường hợp

*) TH1: 3x-5=x+2

=>3x-x=2+5

=>2x=7

=>x=7:2\(\Rightarrow x=\frac{7}{2}\)

*)TH2: -3x+5=x+2

\(\Rightarrow5-3x=x+2\)

\(\Rightarrow5-2=x+3x\)

\(\Rightarrow3=4x\)

\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)

Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)

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

Tìm x biết:

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

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

\(\Rightarrow\frac{x}{2018}+1+\frac{x+1}{2017}+1+\frac{x+2}{2016}+1+\frac{x+3}{2015}+1=0\)

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

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

\(\Rightarrow x+2018=0\) vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}>0\)

\(\Rightarrow x=-2018\)

Vậy x=-2018 thỏa mãn

x2018 +x+12017 +x+22016 +x+32015 =−4

⇒x2018 +1+x+12017 +1+x+22016 +1+x+32015 +1=0

⇒x+20182018 +x+20182017 +x+20182016 +x+20182015 =0

⇒(x+2018)(12018 +12017 +12016 +12015 )=0

⇒x+2018=0 vì 12018 +12017 +12016 +12015 >0

⇒x=−2018

Vậy x=-2018 thỏa mãn

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

trừ mỗi vế cho 2 rồi tách -2 thành -1và -1

X=1 nhé

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

\(\Leftrightarrow\)\(\frac{x+2014}{2015}-1+\frac{x+2015}{2016}-1=\frac{x+2016}{2017}-1+\frac{x+2017}{2018}-1\)

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

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

\(\Leftrightarrow\)\(x-1=0\)   ( do  1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )

\(\Leftrightarrow\) \(x=1\)

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

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

\(\Rightarrow\orbr{\begin{cases}x=2019\left(1\right)\\\frac{1}{2018}+\frac{1}{2017}=\frac{1}{2016}+\frac{1}{2015}\left(2\right)\end{cases}}\)  mà \(\left(2\right)\)không thể xảy ra nên x=2019 là nghiệm của phương trình.

Em chuyển sang cùng một vế rồi ghép cái đầu vói cái thứ 3 cái thứ 2 với cái cuối. :)Dùng quy đồng :)

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

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

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)

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

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

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy : \(x=-2020\)

Chúc bạn học tốt !!

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)

Vậy x = -2020

b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

Vậy x = -2010