3:
1: Thay x=3+2căn 2 vào B, ta được:
\(B=\dfrac{3+2\sqrt{2}+12}{\sqrt{2}+1-1}=\dfrac{15+2\sqrt{2}}{\sqrt{2}}=\dfrac{15\sqrt{2}+4}{2}\)
2:
\(A=\dfrac{\sqrt{x}-2-4\sqrt{x}-8+x+12}{x-4}=\dfrac{x-3\sqrt{x}+2}{x-4}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-1\right)}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\cdot\dfrac{x+2}{\sqrt{x}-1}=\dfrac{x+2}{\sqrt{x}+2}\)
\(=\dfrac{x-4+6}{\sqrt{x}+2}\)
\(=\sqrt{x}-2+\dfrac{6}{\sqrt{x}+2}\)
\(=\sqrt{x}+2+\dfrac{6}{\sqrt{x}+2}-4\)
=>\(P>=2\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{6}{\sqrt{x}+2}}-4=2\sqrt{6}=-4\)
Dấu = xảy ra khi (căn x+2)^2=6
=>căn x+2=căn 6
=>căn x=căn 6-2
=>x=10-4*căn 6
