a: \(P=\dfrac{x\sqrt{x}+5\sqrt{x}-12}{x-\sqrt{x}-6}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+2}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}+5\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+2}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{x\sqrt{x}+5\sqrt{x}-12-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+5\sqrt{x}-12-2\left(x-6\sqrt{x}+9\right)-x-5\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}-x-18-2x+12\sqrt{x}-18}{\left(\sqrt[]{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}-3x+12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+12\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+12}{\sqrt{x}+2}\)
b: \(P=\dfrac{x+12}{\sqrt{x}+2}=\dfrac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{16}{\sqrt{x}+2}\)
=>\(P=\sqrt{x}+2+\dfrac{16}{\sqrt{x}+2}-4\)
=>\(P>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{16}{\sqrt{x}+2}}-4=2\cdot4-4=4\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\left(\sqrt{x}+2\right)^2=16\)
=>\(\sqrt{x}+2=4\)
=>x=4(nhận)
