c) <=> \(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1\)\(+\frac{x+3}{2014}+1\)= \(\frac{x+4}{2013}+1+\frac{x+5}{2012}+1\)\(+\frac{x+6}{2011}\)
<=> \(\frac{x+1+2016}{2016}+\frac{x+2+2015}{2015}+\frac{x+3+2014}{2014}\) \(=\frac{x+4+2013}{2013}+\frac{x+5+2012}{2012}+\frac{x+6+2011}{2011}\)
<=> \(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
<=> \(\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
vì \(\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)khác 0
=> \(x+2017=0\) => \(x=-2017\)
Vậy \(S=\left\{-2017\right\}\)
a) \(x^2-8x+21=0\)
\(\Leftrightarrow x^2-8x+16+5=0\)
\(\Leftrightarrow\left(x-4\right)^2+5=0\)
Mà \(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2+5>0\)
Vậy pt vô nghiệm
b) \(x^2-5x+9=0\)
\(\Leftrightarrow x^2-2.\frac{5}{2}x+\frac{25}{4}+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2+\frac{11}{4}=0\)
Vì \(\left(x-\frac{5}{2}\right)^2\ge0\Rightarrow\)\(\left(x-\frac{5}{2}\right)^2+\frac{11}{4}>0\)
Vậy pt vô nghiệm
