`1)` Với `x >= 0,x \ne 4` có:
`P=[x+\sqrt{x}]/[\sqrt{x}-2]-[2\sqrt{x}-1]/[\sqrt{x}+2]+[x-6\sqrt{x}+4]/[x-4]`
`P=[(x+\sqrt{x})(\sqrt{x}+2)-(2\sqrt{x}-1)(\sqrt{x}-2)+x-6\sqrt{x}+4]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`P=[x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`P=[x\sqrt{x}+2x+\sqrt{x}+2]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`P=[x(\sqrt{x}+2)+(\sqrt{x}+2)]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`P=[(\sqrt{x}+2)(x+1)]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`P=[x+1]/[\sqrt{x}-2]`
__________________________________________________
`2)x=9+4\sqrt{5}` t/m đk
`=>\sqrt{x}=\sqrt{9+4\sqrt{5}}=\sqrt{(\sqrt{5}+2)^2}=\sqrt{5}+2`
Thay `x=9+4\sqrt{5}` và `\sqrt{x}=\sqrt{5}+2` vào `P` có:
`P=[9+4\sqrt{5}]/[\sqrt{5}+2-2]`
`P=[9+4\sqrt{5}]/\sqrt{5}`
`P=[\sqrt{5}(9+4\sqrt{5})]/5=[9\sqrt{5}+20]/5`
đk x ≠4 ; x ≥ 0
\(P=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{x-4}\\ =\dfrac{x\sqrt{x}+x+2x+2\sqrt{x}-\left(2x-\sqrt{x}-4\sqrt{x}+2\right)+x-6\sqrt{x}+4}{x-4}\\ =\dfrac{3x+x\sqrt{x}+2\sqrt{x}-2x+5\sqrt{x}-2+x-6\sqrt{x}+4}{x-4}\\ =\dfrac{2x-\sqrt{x}+x\sqrt{x}+2}{x-4}\\=\dfrac{2\left(x-1\right)+\sqrt{x}\left(x-1\right)}{x-4}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{x-4}\\ =\dfrac{x-1}{\sqrt{x}-2}\)
=\(\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{x-4}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}+\dfrac{x-6\sqrt{x}+4}{x-4}\)
=\(\dfrac{x\sqrt{x}+2x+2+\sqrt{x}}{x-4}=\dfrac{\left(\sqrt{x}+2\right)\left(x+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+1}{\sqrt{x}-2}\)
1, ĐKXĐ : x ≥ 0; x ≠ 4
\(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x+4}}{x-4}\)
\(P=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x-1}\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{x\sqrt{x}+3x+2\sqrt{x}-\left(2x-5\sqrt{x}+2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x-2}\right)\left(\sqrt{x+2}\right)}\)
\(P=\dfrac{x\sqrt{x}+2x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)
\(P=\dfrac{x\left(\sqrt{x}+2\right)+\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x+2}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+1}{\sqrt{x}-2}\)
vậy \(P=\dfrac{x+1}{\sqrt{x}-2}\)
2,\(x=9+4\sqrt{5}=5+2.\sqrt{5}.2+4=\left(\sqrt{5}+2\right)^2\)
⇒\(\sqrt{x}=\sqrt{\left(5+2\right)^2}=\sqrt{5}+2\)
⇒\(P=\dfrac{9+4\sqrt{5}+1}{\sqrt{5}+2-2}=\dfrac{10+4\sqrt{5}}{\sqrt{5}}=2\sqrt{5}+4\)
