Câu 3:
a) \(pt\Leftrightarrow\left|2x-3\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=10\\2x-3=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\left(tm\right)\end{matrix}\right.\)
Câu 4:
ĐKXĐ: \(x\ge0\)
\(x-\sqrt{x}+1=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{x-\sqrt{x}+1}\le\dfrac{1}{\dfrac{1}{4}}=4\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
