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19 tháng 6 2017

Ta có : \(\frac{x-1}{2017}+\frac{x-2}{2018}-\frac{x-3}{2019}=\frac{x-4}{2020}\)

\(\Rightarrow\frac{x-1}{2017}+\frac{x-2}{2018}=\frac{x-4}{2020}+\frac{x-3}{2019}\)

\(\Rightarrow1+\frac{x-1}{2017}+1+\frac{x-2}{2018}=1+\frac{x-4}{2020}+1+\frac{x-3}{2019}\)

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

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

\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\text{Mà : }\)\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)

\(\text{Nên : }\) \(2016+x=0\)

\(\Rightarrow x=-2016\)

1 tháng 1 2018

Giỏi wá!!!!!!!!

Ta có:\(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

Mà \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}>0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

\(S=\left\{-2015\right\}\)

16 tháng 4 2020

gợi ý 

2017-x-2=2018-3-x=2019-4-x=2020-5-x

5 tháng 4 2020

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:..............

30 tháng 6 2017

Đề là gì vậy bạn !

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 ~ 

28 tháng 8 2019

a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)

<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)

<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)

=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))

<=> x=-1

Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)

b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)

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

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

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

=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))

<=> x=-2021

Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)

c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)

<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)

<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)

<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)

<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))

<=> x=2010

Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)

d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)

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

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

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

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

<=> x=100

Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)

28 tháng 8 2019

a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)

\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1.\)

Mình chỉ làm câu a) thôi nhé.

Chúc bạn học tốt!

5 tháng 1 2020

\(\frac{x-4}{2017}+\frac{x-3}{2018}+\frac{x-2}{2019}+\frac{x-1}{2020}=4\\ \Leftrightarrow\left(\frac{x-4}{2017}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-2}{2019}-1\right)+\left(\frac{x-1}{2020}-1\right)=4-1-1-1\)

\(\Leftrightarrow\frac{x-2021}{2017}+\frac{x-2021}{2018}+\frac{x-2021}{2019}+\frac{x-2021}{2020}=0\)

\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2021=0\\\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\ne0\end{matrix}\right.\)

\(\Leftrightarrow x=2021\)

Vậy...