Câu 1:
`1, A=(2\sqrt{x})/(\sqrt{x}+3)` (ĐK:`x>=0`)
Thay `x=36` vào A ta có:
`A=(2\sqrt{36})/(\sqrt{36}+3)=(2.6)/(6+3)=12/9=4/3`
Vậy `A=4/3` khi `x=36`
`2, B=(\sqrt{x}+1)/(\sqrt{x}-3)+(11\sqrt{x}-3)(x-9)` (ĐK: `x>=0; x\ne9`)
`B=((\sqrt{x}+1)(\sqrt{x}+3)+11\sqrt{x}-3)/((\sqrt{x}-3)(\sqrt{x}+3))`
`B=(x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3)/((\sqrt{x}-3)(\sqrt{x}+3))`
`B=(x+15\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}+3))`
`-> P=A+B`
`P=(2\sqrt{x}(\sqrt{x}-3)+x+15\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}+3))`
`P=(2x-6\sqrt{x}+x+15\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}+3))`
`P=(3x+9\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}+3))`
`P=(3\sqrt{x}(\sqrt{x}+3))/((\sqrt{x}-3)(\sqrt{x}+3))`
`P=(3\sqrt{x})/(\sqrt{x}-3)`
Vậy `P=(3\sqrt{x})/(\sqrt{x}-3)` với `x>=0; x\ne9`
`3, P=P^4`
`<=> P^4-P=0`
`<=> P(P^3-1)=0`
`<=> P(P-1)(P^2+P+1)=0`
Do `P^2+P+1=(P+1/2)^2+3/4>0` với `AAP`
`->`\(\left[ \begin{array}{l}P=0\\P=1\end{array} \right.\)
* `P=0-> (3\sqrt{x})/(\sqrt{x}-3)=0`
`-> 3\sqrt{x}=0->x=0` (tm)
* `P=1-> 3\sqrt{x}=\sqrt{x}-3`
`<=> 2\sqrt{x}=-3` (vô lý)
Vậy `x=0` thì `P=P^4`
