Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)
\(\Leftrightarrow\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-3}{2018}-1\)
\(\Leftrightarrow\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
\(\Leftrightarrow x-2020=0\)
\(\Leftrightarrow x=2020\)
\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)
\(< =>\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-2}{2018}-1\)
\(< =>\frac{x-5-2015}{2015}+\frac{x-4-2016}{2016}=\frac{x-3-2017}{2017}+\frac{x-2-2018}{2018}\)
\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)
\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}-\frac{x-2020}{2017}-\frac{x-2020}{2018}=0\)
\(< =>\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)
\(< =>x-2020=0< =>x=2020\)
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
Ta có: \(\frac{x-5}{2016}+\frac{x-4}{2017}=\frac{x-3}{2018}+\frac{x-2}{2019}\)
\(\Leftrightarrow\frac{x-5}{2016}-1+\frac{x-4}{2017}-1=\frac{x-3}{2018}-1+\frac{x-2}{2019}-1\)
\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}=\frac{x-2021}{2018}+\frac{x-2021}{2019}\)
\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}-\frac{x-2021}{2018}-\frac{x-2021}{2019}=0\)
\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
mà \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\ne0\)
nên x-2021=0
hay x=2021
Vậy: x=2021
Cộng 2 vế của phương trình với 2 ta có: \(\frac{2-x}{2016}+1=\left(\frac{1-x}{2017}+1\right)-\left(\frac{x}{2018}-1\right)\)
\(\Leftrightarrow\frac{2018-x}{2016}=\frac{2018-x}{2017}-\frac{x-2018}{2018}\)\(\Leftrightarrow\frac{2018-x}{2016}=\frac{2018-x}{2017}+\frac{2018-x}{2018}\)
\(\Leftrightarrow\frac{2018-x}{2016}-\frac{2018-x}{2017}-\frac{2018-x}{2018}=0\)\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Vì \(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)\(\Rightarrow2018-x=0\)\(\Leftrightarrow x=2018\)
Vậy tập nghiệm của phương trình là \(S=\left\{2018\right\}\)
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
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 \(S=\left\{-1;2\right\}\)
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
mk ko chép lại đề nha:
\(\Rightarrow\)\(\frac{x-2}{2017}\)\(-1+\frac{x-3}{2016}\)\(-1=\frac{x-4}{2015}\)\(-1+\frac{x-5}{2014}\)\(-1\)
\(\Rightarrow\)\(\frac{x-2-2017}{2017}\)\(+\frac{x-3-2016}{2016}\)\(=\frac{x-4-2015}{2015}\)\(+\frac{x-5-2014}{2014}\)
\(\Rightarrow\)\(\frac{x-2019}{2017}\)\(+\frac{x-2019}{2016}\)\(-\frac{x-2019}{2015}\)\(-\frac{x-2019}{2014}\)\(=0\)
\(\Rightarrow\)\(\left(x-2019\right)\)\(\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)\)\(=0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-2019=0\\\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}=0\left(voli\right)\end{cases}}\)
\(\Rightarrow\)\(x-2019=0\)
\(\Rightarrow\)\(x=-2019\)
Chỗ mình nghi voli là vô lí nha
chúc bạn học tốt
Ta có \(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}.x=\frac{2018}{2019}.x\)
<=>\(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}x-\frac{2018}{2019}x=0\)
<=>x\(\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
Vì \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\) không thể bằng 0
Vậy x=0
Ta có 1 nghiệm thỏa mãn S=\(\left\{0\right\}\)