Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
d, 2x2-5x-3 = 0
\(\Leftrightarrow\)2x2-6x+x-3= 0
\(\Leftrightarrow\)(2x2-6x) +(x-3) = 0
\(\Leftrightarrow\)2x(x-3) + (x-3) = 0
\(\Leftrightarrow\)(x-3) (2x+1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S =\(\left\{3;\frac{-1}{2}\right\}\)
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:..............
Ta có : \(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}+\frac{x+2038}{6}=0\)
=> \(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1+\frac{x+2038}{6}-3=0\)
=> \(\frac{x+2}{2018}+\frac{2018}{2018}+\frac{x+3}{2017}+\frac{2017}{2017}+\frac{x+4}{2016}+\frac{2016}{2016}+\frac{x+2038}{6}-\frac{18}{6}=0\)
=> \(\frac{x+2000}{2018}+\frac{x+2000}{2017}+\frac{x+2000}{2016}+\frac{x+2000}{6}=0\)
=> \(\left(x+2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\right)=0\)
=> \(x+2000=0\)
=> \(x=-2000\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{-2000\right\}\)
\(\frac{x-1}{2018}+\frac{x-2}{2017}+\frac{x-3}{2016}+\frac{x-2043}{8}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x-1}{2018}-1+\frac{x-2}{2017}-1+\frac{x-3}{2016}-1\)\(+\frac{x-2043}{8}+3=0\)
\(\Leftrightarrow\)\(\frac{x-1}{2018}-\frac{2018}{2018}+\frac{x-2}{2017}-\frac{2017}{2017}\)\(+\frac{x-3}{2016}-\frac{2016}{2016}+\frac{x-2043}{8}+\frac{24}{8}=0\)
\(\Leftrightarrow\)\(\frac{x-2019}{2018}+\frac{x-2019}{2017}+\frac{x-2019}{2016}\)\(+\frac{x-2019}{8}=0\)
\(\Leftrightarrow\)\(\left(x-2019\right).\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow\)\(x-2019=0\) ( Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{8}\ne0\))
\(\Leftrightarrow\) \(x=2019\)
Vậy phương trình có nghiệm là : \(x=2019\)
Giải các pt sau:
a) (x+4)(2x-3)=0
TH1: x+4=0 => x=-4
TH2 : 2x-3=0 => 2x=3 =>x=3/2
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
\(\frac{x-2}{2016}+\frac{x-3}{2017}+\frac{x-4}{2018}+3=0\\ \Leftrightarrow\left(\frac{x-2}{2016}+1\right)+\left(\frac{x-3}{2017}+1\right)+\left(\frac{x-4}{2018}+1\right)=0\\ \Leftrightarrow\frac{x+2014}{2016}+\frac{x+2014}{2017}+\frac{x+2014}{2018}=0\\ \Leftrightarrow\left(x+2014\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\\ Vì\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\ne0\\ \Rightarrow x+2014=0\\ \Leftrightarrow x=-2014\\ Vậy...\)
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 \(S=\left\{-1;2\right\}\)