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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 \(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}.x=\frac{2018}{2019}.x\)
<=>\(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}x-\frac{2018}{2019}x=0\)
<=>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=\(\left\{0\right\}\)
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:\(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Mà \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}>0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
\(S=\left\{-2015\right\}\)
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 : \(x=2019\)
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
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)
\(\Leftrightarrow\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-3}{2018}-1\)
\(\Leftrightarrow\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
\(\Leftrightarrow x-2020=0\)
\(\Leftrightarrow x=2020\)
\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)
\(< =>\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-2}{2018}-1\)
\(< =>\frac{x-5-2015}{2015}+\frac{x-4-2016}{2016}=\frac{x-3-2017}{2017}+\frac{x-2-2018}{2018}\)
\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)
\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}-\frac{x-2020}{2017}-\frac{x-2020}{2018}=0\)
\(< =>\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)
\(< =>x-2020=0< =>x=2020\)