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29 tháng 2 2020

Ta có : \(\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}+\frac{x+5}{1008}+\frac{x+2076}{11}=0\)

=> \(\frac{x+2}{2019}+1+\frac{x+3}{2018}+1+\frac{x+4}{2017}+1+\frac{x+5}{1008}+2+\frac{x+2076}{11}-5=0\)

=> \(\frac{x+2}{2019}+\frac{2019}{2019}+\frac{x+3}{2018}+\frac{2018}{2018}+\frac{x+4}{2017}+\frac{2017}{2017}+\frac{x+5}{1008}+\frac{2016}{1008}+\frac{x+2076}{11}-\frac{55}{11}=0\)

=> \(\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}+\frac{x+2021}{1008}+\frac{x+2021}{11}=0\)

=> \(\left(x+2021\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{1008}+\frac{1}{11}\right)=0\)

=> \(x+2021=0\)

=> \(x=-2021\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{-2021\right\}\)

29 tháng 2 2020

cảm ơn bạn

4 tháng 9 2019

đề thiếu hả bn

4 tháng 9 2019

Hình như bạn ghi thiếu đề rồi. Để tìm đc x trong đẳng thức này thì ta phải có kết quả của biểu thức trên chứ đề cộc lốc thế này ko giải đc đâu

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!!!!!!!!

2 tháng 8 2019

a) \(\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

\(\Leftrightarrow\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)

\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}>0\)

\(\Rightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

b) \(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)

\(\Leftrightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)

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

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

Vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)

\(\Rightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

2 tháng 8 2019

a) \(\left(x+2\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

=>\(x+2=0\)

=>\(x=-2\)

nếu có sai thì mong bn thông cảm nha

12 tháng 1 2020

\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}-1\right)+\left(\frac{x+2}{2018}-1\right)=\left(\frac{x+3}{2017}-1\right)+\left(\frac{x+4}{2016}-1\right)\)

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

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

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

\(\Leftrightarrow x+2020=0\)

Còn lại tự làm :V

12 tháng 1 2020

Lộn chỗ này , thay chút nha ! 

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)

Sorry =))

17 tháng 8 2019

\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)

\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)

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

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

\(\Leftrightarrow\left(x-2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

\(\Leftrightarrow x-2020=0\)

\(\Leftrightarrow x=0+2020\)

\(\Rightarrow x=2020\)

Vậy \(x=2020.\)

Chúc bạn học tốt!

16 tháng 7 2018

<=>[ (x-1)/2019] -1 +[(x-2)/2018]-1 = [(x-3)/2017]-1 +[(x-4)/2016] -1

<=> (x-2020)/2019 +(x-2020)/2018 = (x-2020)/2017 + (x-2020)/2016

<=> (x-2020)( 1/2019+1/2018-1/2017-1/2016)= 0

=> x-2020= 0 => x= 2020

23 tháng 11 2019

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

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=0\)

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

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

\(\Leftrightarrow x+2020=0\)( vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}>0\) )

\(\Leftrightarrow x=-2020\)

Vậy ...

28 tháng 7 2019

\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)

\(\Leftrightarrow\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}-\frac{x-4}{2016}=0\)

\(\Leftrightarrow\frac{x-1}{2019}-1+\frac{x-2}{2018}-1-\frac{x-3}{2017}+1-\frac{x-4}{2016}+1=0\)

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

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

\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)

28 tháng 7 2019

\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)

\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)

\(\frac{x-1}{2019}+\frac{x-2}{2018}-2=\frac{x-3}{2017}+\frac{x-4}{2016}-2\)

\(\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)

\(\frac{x-1-2019}{2019}+\frac{x-2-2018}{2018}=\frac{x-3-2017}{2017}+\frac{x-4-2016}{2016}\)

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

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

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

\(\Rightarrow x-2020=0\)

Vậy \(x=2020\)