Đáp án:
Giải thích các bước giải:
a) $(4x+3)^2=4(x-1)^2\Leftrightarrow 16x^2+24x+9=4(x^2-2x+1)\\
\Leftrightarrow 16x^2-4x^2+24x+8x-4+9=0\\
\Leftrightarrow 12x^2+32x+5=0\Leftrightarrow 12x^2+30x+2x+5=0\\\Leftrightarrow 6x(2x+5)+(2x+5)=0\Leftrightarrow (6x+1)(2x+5)=0\\ \Leftrightarrow 6x+1=0, 2x+5=0 \Leftrightarrow x=\frac{-1}{6},x=\frac{-5}{2}$
b) $\frac{x}{2012} + \frac{x+1 }{2013} + \frac{x+2}{2014} + \frac{x+3}{2015 }+ \frac{x+4}{2016}= 5\\
\Leftrightarrow (\frac{x}{2012}-1)+(\frac{x+1 }{2013} -1)+(\frac{x+2}{2014}-1)+(\frac{x+3}{2015 }-1)+(\frac{x+4}{2016 }-1)=0\\
\Leftrightarrow \frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\\
\Leftrightarrow (x-2012)(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015})=0\\
\Leftrightarrow x-2012=0\Leftrightarrow x=2012(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\neq 0)$
c) $ \frac{x+6}{1005} + \frac{x+132}{471}+ \frac{x+1008}{168} = -12\\
\Leftrightarrow (\frac{x+6}{1005}+2)+(\frac{x+132}{471}+4)+(\frac{x+1008}{168} +6)=0\\
\Leftrightarrow \frac{x+2016}{1005}+\frac{x+2016}{471}+\frac{x+2016}{168}=0\\
\Leftrightarrow (x+2016)(\frac{1}{1005}+\frac{1}{471}+\frac{1}{168})=0\\
\Leftrightarrow x+2016=0 (\frac{1}{1005}+\frac{1}{471}+\frac{1}{168})\neq 0)\\
\Leftrightarrow x=-2016$
