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

<=>\(\frac{x-2020}{2015}+\frac{x-2020}{2016}-\frac{x-2020}{2017}-\frac{x-2020}{2018}=0\)

<=> \((x-2020)(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018})=0\)

<=>\(x-2020=0\)

<=> \(x=2020\)

Vậy_

b, tương tự

Gợi ý :

Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)

Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)

Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)

bài 3

\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)

=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)

=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

=> x=100

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

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

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

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

\(\Leftrightarrow x-2020=0\)

\(\Leftrightarrow x=2020\)

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

\(< =>\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-2}{2018}-1\)

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

\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)

\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}-\frac{x-2020}{2017}-\frac{x-2020}{2018}=0\)

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

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

\(< =>x-2020=0< =>x=2020\)

\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

a, Làm

\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)

<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)

<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

<=> x+2021=0

<=> x=-2021

Kl:......................

b, Làmmmmm

\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)

<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)

<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)

<=> x=2006

Kl:..............

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

<=> \(\frac{x+2019}{2015}+\frac{x+2019}{2016}=\frac{x+2019}{2017}+\frac{x+2019}{2018}\)

<=>(x+2019).(\(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}>0\)

Vì (\(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}>0\)

=> x+2019>0

=>x>-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...

a) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)

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

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

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

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

\(\Leftrightarrow2018-x=0\) ( vì \(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\))

\(\Leftrightarrow x=2018\)

Vậy nghiệm của pt x=2018

b)\(\frac{x-19}{1999}+\frac{x-23}{1995}+\frac{x+82}{700}=5\)

\(\Leftrightarrow\left(\frac{x-19}{1999}-1\right)+\left(\frac{x-23}{1995}+-1\right)+\left(\frac{x+82}{700}-3\right)=0\)

\(\Leftrightarrow\frac{x-2018}{1999}+\frac{x-2018}{1995}+\frac{x-2018}{700}=0\)

\(\Leftrightarrow\left(x-2018\right)\left(\frac{1}{1999}+\frac{1}{1995}+\frac{1}{700}\right)=0\)

\(\Leftrightarrow x-2018=0\)( vì \(\frac{1}{1999}+\frac{1}{1995}+\frac{1}{700}\ne0\))

\(\Leftrightarrow x=2018\)

Vậy nghiệm của pt x=2018

c) \(x^3-3x^2+4=0\)

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

\(\Leftrightarrow x^2\left(x+1\right)-4\left(x^2-1\right)=0\)

\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

Vậy tập hợp nghiệm \(S=\left\{-1;2\right\}\)