Bài 2:
a) \(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)+6\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x^2+x-2=0\) hay \(x^2+x+6=0\)
\(\Leftrightarrow x^2-x+2x-2=0\) hay \(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{23}{4}=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\) hay \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\) (pt vô nghiệm)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=1\) hay \(x=-2\)
-Vậy \(S=\left\{1;-2\right\}\)
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\Leftrightarrow\)\(\left(x^2+x\right)^2+4\left(x^2+x\right)+4=16\)
\(\Leftrightarrow\left(x^2+x+2\right)^2=16\)
\(\Leftrightarrow\left(x^2+x+2\right)^2-16=0\)
\(\Leftrightarrow\left(x^2+x+2+4\right)\left(x^2+x+2-4\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=> \(x^2+x+6=0\) hoặc \(x^2+x-2=0\)
\(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=0\) \(x^2+2x-x-2=0\)
\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\left(vô lý\right)\) \(x\left(x+2\right)-\left(x+2\right)=0\)
\(\left(x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy....
b)
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
\(\Leftrightarrow\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
\(\Leftrightarrow\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
\(\Leftrightarrow\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\ne0\)
=> x+2009=0
x=-2009
Vậy....
b) \(\Leftrightarrow\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)\(\Leftrightarrow\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)
-Vì \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)< 0\) nên pt đã cho tương đương:
\(x+2009=0\Leftrightarrow x=-2009\)
Vậy \(S=\left\{-2009\right\}\)
