Bài 1
$a)\ (4-5x)(x+2)\ge0\\\Leftrightarrow \left[\begin{matrix}\left\{\begin{matrix}4-5x\ge0\\x+2\ge0\end{matrix}\right.\\\left\{\begin{matrix}4-5x\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}\left\{\begin{matrix}x\le\dfrac{4}{5}\\x\le-2\end{matrix}\right.\\\left\{\begin{matrix}x\ge\dfrac{4}{5}\\x\le-2\end{matrix}\right.\end{matrix}\right.$
$\Leftrightarrow -2\le x\le \dfrac{4}{5}$
Vậy...
$b)x^2-7\ge0\\\Leftrightarrow x^2\ge7\\\Leftrightarrow \left[\begin{matrix}x\le-\sqrt{7}\\x\ge\sqrt{7}\end{matrix}\right.$
Vậy...
$c)(11-3x)(4-7x)\ge0\\\Leftrightarrow \left[\begin{matrix}\left\{\begin{matrix}11-3x\ge0\\4-7x\ge0\end{matrix}\right.\\\left\{\begin{matrix}11-3x\le0\\4-7x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}\left\{\begin{matrix}x\le\dfrac{11}{3}\\x\le\dfrac{4}{7}\end{matrix}\right.\\\left\{\begin{matrix}x\ge\dfrac{11}{3}\\x\ge\dfrac{4}{7}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}x\le\dfrac{4}{7}\\x\ge\dfrac{11}{3}\end{matrix}\right.$
Vậy...
$d)x^2+7x+6\ge0\\\Leftrightarrow (x+1)(x+6)\ge0\\\Leftrightarrow \left[\begin{matrix}x\ge-1\\x\le-6\end{matrix}\right.$
Vậy...
Bài 2
$a)\sqrt{x^2-4x+4}=5\\\Leftrightarrow \sqrt{(x-2)^2}=5\\\Leftrightarrow |x-2|=5\\\Leftrightarrow \left[\begin{matrix}x=7\\x=-3\end{matrix}\right.$
$b)Đk: 2x-5\ge0\Leftrightarrow x\ge\dfrac{5}{2}$
$\sqrt{9x^2+6x+1}=2x-5\\\Leftrightarrow\sqrt{(3x+1)^2}=2x-5\\\Leftrightarrow |3x+1|=2x-5\\\Leftrightarrow \left[\begin{matrix}3x+1=2x-5\\3x+1=5-2x\end{matrix}\right.$
$\Leftrightarrow\left[\begin{matrix}x=-6\\x=\dfrac{4}{5}\end{matrix}\right.(loại)$
Vậy pt vô nghiệm
$c)\sqrt{x^2+2\sqrt{2}x+2}=1\\\Leftrightarrow \sqrt{(x+\sqrt{2})^2}=1\\\Leftrightarrow |x+\sqrt{2}|=1\\\Leftrightarrow \left[\begin{matrix}x=1-\sqrt{2}\\x=-1-\sqrt{2}\end{matrix}\right.$
Vậy...
$d)\sqrt{x^2-4x+4}-3x+2=0\\\Leftrightarrow \sqrt{(x-2)^2}=3x-2\ (Đk:x\ge\dfrac{2}{3})\\\Leftrightarrow |x-2|=3x-2\\\Leftrightarrow \left[\begin{matrix}x-2=3x-2\\x-2=2-3x\end{matrix}\right.$
$\Leftrightarrow \left[\begin{matrix}x=0\ (loại)\\x=1\ (t/m)\end{matrix}\right.$
Vậy pt có nghiệm là $x=1$
