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

\(PT\Leftrightarrow\left(\frac{x-5}{2020}-1\right)+\left(\frac{x-6}{2019}-1\right)-\left(\frac{x-7}{2018}-1\right)-\left(\frac{x-8}{2017}-1\right)=0\)

\(\Leftrightarrow\left(x-2025\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

Dễ thấy \(\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)< 0\)

\(\Rightarrow x=2025=5^2.3^4\)

Vậy các ước nguyên tố của nghieemh pt là 3,5 

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

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

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

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

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

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

\(\text{Nên : }\) \(2016+x=0\)

\(\Rightarrow x=-2016\)

Giỏi wá!!!!!!!!

Ta có : x−12017 +x−22018 −x−32019 =x−42020 

⇒x−12017 +x−22018 =x−42020 +x−32019 

⇒1+x−12017 +1+x−22018 =1+x−42020 +1+x−32019 

⇒2016+x2017 +2016+x2018 =2016+x2020 +2016+x2019 

⇒2016+x2017 +2016+x2018 −2016+x2019 −2016+x2020 =0

⇒(2016+x)(12017 +12018 −12019 −12020 )=0
Mà : 12017 +12018 −12019 −12020 ≠0

Nên : 2016+x=0

⇒x=−2016

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

\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}+\frac{x+2036}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1+\frac{x+2038}{6}-3=0\)

\(\Leftrightarrow\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}+\frac{x+2020}{6}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\right)=0\)

có : \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\ne0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

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

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

\(\Leftrightarrow\frac{x-2021}{2018}+\frac{x-2021}{2019}=\frac{x-2021}{2}+\frac{x-2021}{3}\)

bài 3 thì lần lượt trừ đi 1; 2; 3; 4; 5

Đề là gì vậy bạn !

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!

Lê Thị Thục HiềnTrần Thanh PhươngVũ Minh Tuấn?Amanda?Nguyễn Việt LâmHISINOMA KINIMADONguyễn Huy TúAkai Haruma

buithianhtho làm cách này mà ko có máy tính thì đến bao giờ ?

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

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

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

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

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

Vì \(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\ne0\)

\(\Leftrightarrow x-2020=0\)

\(\Leftrightarrow x=2020\)

Vậy....

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

= 4074342(x-3)+4072323(x-2)+4070306(x-1)=24653843442

=07342x- 12223026+ 4072323x-8144646+4070306x- 4070306= 24653843442

12216971x- 24437978= 24653843442

12216971x=24653843442+24437978

12216971x= 24678281420

x= 2020

chỗ dòng thứ 3 số đầu tiên là 4074342x-....