giải phương trình \(\frac{2017}{x^2+2017}+\frac{2018}{x^2+2018}=2\)
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13 tháng 3 2018
Dễ thấy \(x=2017\)không là nghiệm của phương trình.
Ta có:
\(\frac{1+\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)^2}{1-\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)}=\frac{13}{37}\)
Đặt \(\frac{x-2018}{2017-x}=a\)
\(\Rightarrow\frac{1+a+a^2}{1-a+a^2}=\frac{13}{37}\)
\(\Leftrightarrow24a^2+50a+24=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\\a=-\frac{4}{3}\end{cases}}\)
HY
9 tháng 2 2020
\(\frac{x-3}{2017}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\\\Leftrightarrow \left(\frac{x-3}{2017}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-2018}{2}-1\right)+\left(\frac{x-2017}{3}-1\right)\\\Leftrightarrow \frac{x-2020}{2017}+\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\\ \Leftrightarrow\frac{x-2020}{2017}+\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\\ \Leftrightarrow\left(x-2020\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\right)=0\\ \Leftrightarrow x-2020=0\left(Vi\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\ne0\right)\\ \Leftrightarrow x=2020\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{2020\right\}\)
15 tháng 5 2020
\(\frac{x-3}{2017}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\)
\(\Leftrightarrow\) \(\frac{x-3}{2017}-1+\frac{x-2}{2018}-1=\frac{x-2018}{2}-1+\frac{x-2017}{3}-1\)
\(\Leftrightarrow\) \(\frac{x-2020}{2017}+\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\)
\(\Leftrightarrow\) \(\frac{x-2020}{2017}+\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\)
\(\Leftrightarrow\) (x - 2020)(\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\)) = 0
\(\Leftrightarrow\) x - 2020 = 0
\(\Leftrightarrow\) x = 2020
Vậy S = {2020}
Chúc bn học tốt!!
16 tháng 4 2020
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 \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}>0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
\(S=\left\{-2015\right\}\)
HT
1
S
5 tháng 3 2019
\(\frac{x-3}{2017}-\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\)
\(\Leftrightarrow\frac{x-3}{2017}-1-\frac{x-2}{2018}-1=\frac{x-2018}{2}-1+\frac{x-2017}{3}-1\)
\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\)
\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)
HT
1
MP
29 tháng 6 2020
\(\frac{2-x}{2017}-1=\frac{1-x}{2018}-\frac{x}{2019}\)
\(\Leftrightarrow\) \(\frac{2-x}{2017}+1=\frac{1-x}{2018}+1-\frac{x}{2019}+1\)
\(\Leftrightarrow\) \(\frac{2019-x}{2017}=\frac{2019-x}{2018}-\frac{2019-x}{2019}\)
\(\Leftrightarrow\) \(\frac{2019-x}{2017}-\frac{2019-x}{2018}+\frac{2019-x}{2019}=0\)
\(\Leftrightarrow\) \(\left(2019-x\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\right)=0\)
Mà \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\right)\ne0\)
\(\Rightarrow\) \(2019-x=0\) \(\Leftrightarrow\) \(x=2019\)
\(\Rightarrow\) \(S=\left\{2019\right\}\)
NN
2 tháng 5 2019
\(\frac{x-1}{2018}+\frac{x-2}{2017}+\frac{x-3}{2016}+\frac{x-2043}{8}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x-1}{2018}-1+\frac{x-2}{2017}-1+\frac{x-3}{2016}-1\)\(+\frac{x-2043}{8}+3=0\)
\(\Leftrightarrow\)\(\frac{x-1}{2018}-\frac{2018}{2018}+\frac{x-2}{2017}-\frac{2017}{2017}\)\(+\frac{x-3}{2016}-\frac{2016}{2016}+\frac{x-2043}{8}+\frac{24}{8}=0\)
\(\Leftrightarrow\)\(\frac{x-2019}{2018}+\frac{x-2019}{2017}+\frac{x-2019}{2016}\)\(+\frac{x-2019}{8}=0\)
\(\Leftrightarrow\)\(\left(x-2019\right).\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow\)\(x-2019=0\) ( Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{8}\ne0\))
\(\Leftrightarrow\) \(x=2019\)
Vậy phương trình có nghiệm là : \(x=2019\)
NN
21 tháng 3 2020
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 \(S=\left\{2018\right\}\)
DN
18 tháng 2 2018
À khác cái dấu nhưng đề phải là giải phương trình chứ
Đặt 2017-x=a => x-2018=-a-1 phương trình trở thành:
\(\frac{a^2+a\left(-a-1\right)+\left(a-1\right)^2}{a^2-a\left(-a-1\right)+\left(a-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{3}{2}\\a=-\frac{5}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=2015,5\\x=2019,5\end{cases}}}\)
Vậy......................
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