1, giải bất phương trình
a)x^2-36/4-x <0
b) 3x^2-2x-12 >=0
$a)\dfrac{x^2-36}{4-x} <0
\\\Leftrightarrow \left\{\begin{matrix}x^2-36>0\\ 4-x<0\end{matrix}\right. \ or \ \left\{\begin{matrix}x^2-36<0\\ 4-x>0\end{matrix}\right.
\\\Leftrightarrow \left\{\begin{matrix}x<-6;x>6\\ x>4\end{matrix}\right. \ or \ \left\{\begin{matrix}-6<x<6\\ x<4\end{matrix}\right.
\\\Leftrightarrow x>6 \ or \ -6<x<4$
Vậy...
$b) 3x^2-2x-12\geq 0
\\\Leftrightarrow x^2-\dfrac{2}{3}x-4\geq 0
\\\Leftrightarrow (x^2-\dfrac{2}{3}x+\dfrac{1}{9})-\dfrac{37}{9}\geq 0
\\\Leftrightarrow (x-\dfrac13)^2\geq \dfrac{37}9
\\\Leftrightarrow \left | x-\dfrac13 \right |\geq \dfrac{\sqrt{37}}3
\\\Leftrightarrow x-\dfrac13\leq \dfrac{-\sqrt{37}}{3};x-\dfrac13\geq \dfrac{\sqrt{37}}3
\\\Leftrightarrow x\leq \dfrac{1-\sqrt{37}}{3};x\geq \dfrac{1+\sqrt{37}}{3}$
Vậy...
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