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NV
31 tháng 1 2019

\(\dfrac{2-x}{2017}+1=\dfrac{x-1}{2018}-1+1-\dfrac{x}{2019}\)

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

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

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

\(\Leftrightarrow2019-x=0\) (do \(\dfrac{1}{2017}>\dfrac{1}{2019}\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}>0\))

\(\Rightarrow x=2019\)

26 tháng 1 2019

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

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

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)

Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)

\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)

Vậy...

6 tháng 1 2018

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

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

Ta đã có: \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}< 0\)

Vậy ta dễ dàng suy ra được \(S=\left\{2019\right\}\)

7 tháng 1 2018

Cố gắng học Thư nhé! Say oh yeah!

27 tháng 1 2019

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

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

\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{3}-\dfrac{x+2020}{4}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x+2020=0\) ( do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\))

\(\Leftrightarrow x=-2020\)

Vậy phương trình có tập nghiệm S = \(\left\{-2020\right\}\)

1 tháng 3 2017

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

\(\Leftrightarrow\frac{x-1}{2017}+1+\frac{x-2}{2018}+1+\frac{x-3}{2019}+1=0\)

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

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

\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\ne0\) nên

x+2016=0

\(\Leftrightarrow\)x=-2016

4 tháng 3 2018

a) ĐKXĐ: \(x\ne\pm2\)

Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)

\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow x\left(x-2\right)=x^2+4\)

\(\Leftrightarrow x^2-2x=x^2+4\)

\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)

Vậy phương trình vô nghiệm

4 tháng 3 2018

b) ĐKXĐ: \(x\ne3;x\ne-1\)

Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)

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

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)

\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm là \(x=0\)

10 tháng 3 2018

Sửa đề: \(\dfrac{x-4}{2019}+\dfrac{x-3}{2018}=\dfrac{x-2}{2017}+\dfrac{x-1}{2016}\)

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

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

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

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

20 tháng 3 2018

đoạn cuối cùng mk chưa hiểu lắm leuleu

19 tháng 1 2019

\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)

\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)

Mấy câu khác tương tự :v

b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

=>123-x=0

=>x=123

c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)

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

=>x-2019=0

=>x=2019

13 tháng 3 2022

\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\)x-2-5(x+1)=15

\(\Leftrightarrow\) x-2-5x-5=15

\(\Leftrightarrow\)x-5x=15+2+5

\(\Leftrightarrow\)-4x=22

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

vậy

13 tháng 3 2022

nhớ like nhahaha