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a) \(x^3-6x^2-9x+14=0\)
\(\Leftrightarrow x^3-8x^2+2x^2+7x-16x+14=0\)
\(\Leftrightarrow\left(x^3-8x^2+7x\right)+\left(2x^2-16x+14\right)=0\)
\(\Leftrightarrow x\left(x^2-8x+7\right)+2\left(x^2-8x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-7x-x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x-7\right)-\left(x-7\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow x\in\left\{-2;1;7\right\}\)
a/ Đặt \(x^2+x+1=a\Rightarrow x^2+x+2=a+1\)
Pt trở thành \(a\left(a+1\right)-12=0\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
2/ \(\dfrac{x+1}{2014}+1+\dfrac{x+2}{2013}+1=\dfrac{x+3}{2012}+1+\dfrac{x+4}{2011}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}=\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2015=0\) (do \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
Ta có : | x - 2011 |2011 + | x - 2012 |2012 \(\ge\)0
Mà | x - 2011 |2011 + | x - 2012 |2012 = 1
xét 2 TH :
TH1 : | x - 2011 |2011 = 0 ; | x - 2012 |2012 = 1
\(\Rightarrow\)x = 2011
TH2 : | x - 2011 |2011 = 1 ; | x - 2012 |2012 = 0
\(\Rightarrow\)x = 2012
vậy x = 2011 hoặc x = 2012
+) Xét x < 2011 thì \(x-2012< -1\)
\(\Rightarrow\left|x-2012\right|^{2012}>1\)
Mà \(\left|x-2011\right|^{2011}>0\forall x< 2011\)
\(\Rightarrow VT>1\left(vl\right)\)
+) Xét x = 2011 thì thỏa mãn
+) Xét 2011 < x < 2012 thì \(\hept{\begin{cases}0< x-2011< 1\\-1< x-2012< 0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-2011\right|^{2011}< x-2011\\\left|x-2012\right|^{2012}< 2012-x\end{cases}}\)
\(\Rightarrow VT< 1\left(vl\right)\)
+) Xét x = 2012 thì thỏa mãn
+) Xét x > 2012 thì \(x-2011>1\)
\(\Rightarrow\left|x-2011\right|^{2011}>1\)
và \(\left|x-2012\right|^{2012}>0\forall x>2012\)
\(\Rightarrow VT>1\)(vl)
Vậy tập nghiệm S = {2011;2012}