a: \(A=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{x-9}\)
\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-5}\)
b: P=A:B
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{\sqrt{x}}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(P-1=\dfrac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{-8}{\sqrt{x}+3}< 0\)
=>P<1
d. Ta có:
$G(x) = 6x^4+6x^3+2x^2+1=6(x^4+x^3+\frac{x^2}{4})+\frac{1}{2}x^2+1$
$=6(x^2+\frac{x}{2})^2+\frac{1}{2}x^2+1\geq 6.0+\frac{1}{2}.0+1>0$ với mọi $x$
Do đó $G(x)$ luôn dương với mọi $x$
