Đáp án+Giải thích các bước giải:
$18.\\\sqrt{x^2 +4x} =2\\\Leftrightarrow x^2 +4x =4\\\Leftrightarrow x= \dfrac{-4\pm \sqrt{4^2 -4\times 1 \times (-4)}}{2\times1}\\\Leftrightarrow x= \dfrac{-4\pm\sqrt{32}}{2}\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{-4+4\sqrt{2}}{2}\\\\x=\dfrac{-4-4\sqrt{2}}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-2+2\sqrt{2}\\\\x=-2-2\sqrt{2}\end{array} \right.\\19.\\\sqrt{x^2 -10x +4}=2\\\Leftrightarrow x^2 -10x + 4 =2\\\Leftrightarrow x^2 -10x =0\\\Leftrightarrow x(x-10)=0\\\Leftrightarrow \left[ \begin{array}{l}x=0\\\\x-10=0 \Leftrightarrow x =10\end{array} \right.\\20.\\\sqrt{4x^2 + 4x + 9}=3\\\Leftrightarrow 4x^2 + 4x +9 =9\\\Leftrightarrow 4x^2 + 4x =0\\\Leftrightarrow 4x(x+1)=0\\\Leftrightarrow x(x+1)=0\\\Leftrightarrow \left[ \begin{array}{l}x=0\\\\x+1=0\Leftrightarrow x=-1\end{array} \right.\\21.\\\sqrt{(x-3)^2}=x-3\\\Leftrightarrow |x-3|=x-3\\\Leftrightarrow \left[ \begin{array}{l} x-3-x =-3, x-3 \ge 0\\\\-(x-3)-x =-3, x -3 <0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x \in \mathbb{R}, x \ge 3\\\\x=3, x < 3\\\end{array} \right.\\\Leftrightarrow x\ge 3$
