\(1,B=\dfrac{9\cdot3+5\cdot3}{9-4\cdot3+4}=\dfrac{42}{1}=42\\ 2,\\ a,A=\dfrac{2x+4-x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\left(x>0;x\ne4\right)\\ A=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(x+5\right)}{\left(\sqrt{x}-2\right)^2}=\dfrac{x+5}{\sqrt{x}-2}\\ b,P=\dfrac{x-4+9}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+9}{\sqrt{x}-2}\\ P=\sqrt{x}+2+\dfrac{9}{\sqrt{x}-2}=\left(\sqrt{x}-2\right)+\dfrac{9}{\sqrt{x}-2}+4\\ P\ge2\sqrt{\dfrac{9\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4\left(BĐT.cosi\right)\\ P\ge2\sqrt{9}+4=10\)
Dấu \("="\Leftrightarrow\left(\sqrt{x}-2\right)^2=9\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=3\\\sqrt{x}-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\left(tm\right)\\x\in\varnothing\left(x>0\right)\end{matrix}\right.\Leftrightarrow x=25\)
