1:
a) \(x\ge2\)
b) \(x\ne-\dfrac{1}{2}\)
2)
a) ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow x-3=4\\ \Leftrightarrow x=7\)
b) \(\Leftrightarrow\left|2x+3\right|=7\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=-7\\2x+3=7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
c) ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{x+1}-2\sqrt{x+1}+\dfrac{16}{5}\sqrt{x+1}=11\\ \Leftrightarrow\dfrac{11}{5}\sqrt{x+1}=11\\ \Leftrightarrow\sqrt{x+1}=5\\ \Leftrightarrow x+1=25\\ \Leftrightarrow x=24\)
Bài III
\(a,x=4\Leftrightarrow A=\dfrac{2-3}{4-2+1}=\dfrac{-1}{3}\\ b,B=\dfrac{3\sqrt{x}+6-2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}+3\right)=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c,P=AB=\dfrac{\sqrt{x}-3}{x-\sqrt{x}+1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\\ P=\sqrt{x}-1+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}-1=2-1=1>0\\ \Leftrightarrow\left|P\right|=P\)
