Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a.
\(P=\frac{(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}+\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-1)}-\frac{15\sqrt{x}-11}{(\sqrt{x}+3)(\sqrt{x}-1)}\)
\(=\frac{(2x+\sqrt{x}-3)+(3x+7\sqrt{x}-6)-(15\sqrt{x}-11)}{(\sqrt{x}+3)(\sqrt{x}-1)}=\frac{5x-7\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+3)}\)
\(=\frac{(\sqrt{x}-1)(5\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{5\sqrt{x}-2}{\sqrt{x}+3}\)
b.
\(P=\frac{5(\sqrt{x}+3)-17}{\sqrt{x}+3}=5-\frac{17}{\sqrt{x}+3}\)
Vì $\sqrt{x}\geq 0\Rightarrow \sqrt{x}+3\geq 3$
$\Rightarrow \frac{17}{\sqrt{x}+3}\leq \frac{17}{3}$
$\Rightarrow P=5-\frac{17}{\sqrt{x}+3}\geq 5-\frac{17}{3}=\frac{-2}{3}$
Đây chính là gtnn của $P$ đạt được khi $x=0$
a) \(P=\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-15\sqrt{x}+11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5x-7\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)
b) Ta có: \(P=\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{5\left(\sqrt{x}+3\right)-17}{\sqrt{x}+3}=5-\dfrac{17}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\)
\(\Rightarrow5-\dfrac{17}{\sqrt{x}+3}\ge-\dfrac{2}{3}\Rightarrow P_{min}=-\dfrac{2}{3}\) khi \(x=0\)
a) Ta có: \(P=\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\)
\(=\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-2\sqrt{x}+3\sqrt{x}-3+3x+9\sqrt{x}-2\sqrt{x}-6-15\sqrt{x}+11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{5x-7\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)
