`a)x=36<=>A=(6+1)/(6-8)=-7/2`
`b)B=(sqrtx-1)/(sqrtx+2)-(1-7sqrtx)/(x-sqrtx-6)`
`=((sqrtx-1)(sqrtx-3)+7sqrtx-1)/(x-sqrtx-6)`
`=(x-4sqrtx+3+7sqrtx-1)/(x-sqrtx-6)`
`=(x+3sqrtx+2)/(x-sqrtx-6)`
`=((sqrtx+1)(sqrtx+2))/((sqrtx+2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
`c)P=B/A`
`=(sqrtx+1)/(sqrtx-3).(x-8)/(sqrtx+1)`
`=(x-8)/(sqrtx-3)`
`P<4`
`<=>(x-8)/(sqrtx-3)-4<0`
`<=>(x-8-4sqrtx+12)/(sqrtx-3)<0`
`<=>(x-4sqrtx+4)/(sqrtx-3)<0`
`<=>(sqrtx-2)^2/(sqrtx-3)<0`
Mà `(sqrtx-2)^2>=0`
`<=>sqrtx-3<0<=>x<9,x ne 4`
Vậy `0<=x<=9,x ne 4` thì `P<4`
1, khi x=36 => \(A=\dfrac{\sqrt{36}+1}{36-8}=\dfrac{6+1}{28}=\dfrac{7}{28}=\dfrac{1}{4}\)
2, B=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1-7\sqrt{x}}{x-\sqrt{x}-6}\left(x\ge0,x\ne8,x\ne9\right)\)
=\(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{1-7\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
=\(\dfrac{x-3\sqrt{x}-\sqrt{x}+3-1+7\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
=\(\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)=\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(dpcm\right)\)
3, P=\(\dfrac{B}{A}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}:\dfrac{\sqrt{x}+1}{x-8}=\dfrac{x-8}{\sqrt{x}-3}\)
để P<4 <=>\(\dfrac{x-8}{\sqrt{x}-3}< 4< =>\dfrac{x-8}{\sqrt{x}-3}-4< 0\)
\(< =>\dfrac{x-8-4\sqrt{x}+12}{\sqrt{x}-3}< 0\)
<=>\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-3}< 0< =>\sqrt{x}-3< 0< =>x< 9\)
kết hợp đkxd: =>để P<4 thì \(0\le x< 9\) và x\(\ne9,x\ne8\)
=>x\(\in\left\{0,1,2,3,4,5,6,7\right\}\)
mà x là số nguyên tố nê x\(\in\left\{2;3;5;7\right\}\)
