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Hình như đề sai dấu, mình sửa lại rồi!
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+...+\frac{x-2017}{1}=2017\)
\(\Leftrightarrow\) \(\frac{x-1}{2017}-1+\frac{x-2}{2016}-1+\frac{x-3}{2015}-1+...+\frac{x-2017}{1}-1=0\)
\(\Leftrightarrow\) \(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+...+\frac{x-2018}{1}=0\)
\(\Leftrightarrow\) (x - 2018)\(\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)
\(\Leftrightarrow\) x - 2018 = 0
\(\Leftrightarrow\) x = 2018
Vậy S = {2018}
Chúc bn học tốt!!
Hình như đề sai dấu, mình sửa lại rồi!
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+...+\frac{x-2017}{1}=2017\)
\(\Leftrightarrow\) \(\frac{x-1}{2017}-1+\frac{x-2}{2016}-1+\frac{x-3}{2015}-1+...+\frac{x-2017}{1}-1=0\)
\(\Leftrightarrow\) \(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+...+\frac{x-2018}{1}=0\)
\(\Leftrightarrow\) (x - 2018)\(\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)
\(\Leftrightarrow\) x - 2018 = 0
\(\Leftrightarrow\) x = 2018
Vậy S = {2018}
Chúc bn học tốt!!
Đặt :
\(A=\dfrac{10^{2017}+1}{10^{2016}+1}\)
\(B=\dfrac{10^{2018}+1}{10^{2017}+1}\)
Ta thấy :
\(\left\{{}\begin{matrix}A=\dfrac{10^{2017}+1}{10^{2016}+1}>1\\B=\dfrac{10^{2018}+1}{10^{2017}+1}>1\end{matrix}\right.\)
Áp dung tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\)
\(B=\dfrac{10^{2018}+1}{10^{2017}+1}>\dfrac{10^{2018}+1+9}{10^{2017}+1+9}=\dfrac{10^{2018}+10}{10^{2017}+10}=\dfrac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2017}+1}{10^{2016}+1}=A\) \(\Leftrightarrow B>A\)
Vậy ......................
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=\(\left\{0\right\}\)
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
M > N