Đáp án:
Giải thích các bước giải:
`B=\frac{3\sqrt{x}}{x+2\sqrt{x}-8}-\frac{2}{\sqrt{x}+4}`
`B=\frac{3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+4)}-\frac{2(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+4)}`
`B=\frac{3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+4)}-\frac{2\sqrt{x}-4}{(\sqrt{x}-2)(\sqrt{x}+4)}`
`B=\frac{3\sqrt{x}-2\sqrt{x}+4}{(\sqrt{x}-2)(\sqrt{x}+4)}`
`B=\frac{\sqrt{x}+4}{(\sqrt{x}-2)(\sqrt{x}+4)}`
`B=\frac{1}{\sqrt{x}-2}`
b) Đặt `P=A/B`
`P=\frac{\sqrt{x}+4}{\sqrt{x}-2}:\frac{1}{\sqrt{x}-2}`
`P=\frac{\sqrt{x}+4}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{1}`
`P=\sqrt{x}+4`
`P=x`
`⇔ \sqrt{x}+4=x`
`⇔ x-\sqrt{x}-4=0`
Đặt `\sqrt{x}=t\ (t \ge 0)`
`t^2-t-4=0`
`Δ=(-1)^2-4.1.(-4)=17`
`Δ>0:` PT có 2 nghiệm pb
`t_{1}=\frac{1+\sqrt{17}}{2}\ (TM)`
`t_{2}=\frac{1-\sqrt{17}}{2}\ (L)`
`t=\frac{1+\sqrt{17}}{2}⇔\sqrt{x}=\frac{1+\sqrt{17}}{2}`
`⇔ x=\frac{9+\sqrt{17}}{2}\ (TM)`
Vậy với `x=\frac{9+\sqrt{17}}{2}` thì `A/B=x`
