Đáp án + Giải thích các bước giải:
`a,`
ĐKXĐ : `x>=0 `
`P=((x+2)/(x sqrtx+1)-1/(sqrtx+1)).(4sqrtx)/3`
`=((x+2)/(sqrt(x^3)+1)-1/(sqrtx+1)).(4sqrtx)/3`
`=(x+2-(x-sqrtx+1))/((sqrtx+1)(x-sqrtx+1)).(4sqrtx)/3`
`=(x+2-x+sqrtx-1)/((sqrtx+1)(x-sqrtx+1)).(4sqrtx)/3`
`=(sqrtx+1)/((sqrtx+1)(x-sqrtx+1)).(4sqrtx)/3`
`=1/(x-sqrtx+1) . (4sqrtx)/3`
`=(4sqrtx)/(3(x-sqrtx+1))`
Tại `x=4 qquad text{(tmđk)}` giá trị của `P` là :
`P=(4sqrt4)/(3(4-sqrt4+1))=(4.2)/(3.3)=8/9`
Vậy tại `x=4` thì `P=8/9`
`b,`
`P=8/9 <=> (4sqrtx)/(3x-3sqrtx+3)=8/9`
`<=> 8(3x-3sqrtx+3)=36sqrtx`
`<=> 24x-24sqrtx+24=36sqrtx`
`<=> 24x-24sqrtx-36sqrtx+24=0`
`<=> 24x-60sqrtx+24=0`
`<=> 2x-5sqrtx+2=0`
`<=> 2x-4sqrtx-sqrtx+2=0`
`<=> 2sqrtx(sqrtx-2)-(sqrtx-2)=0`
`<=> (2sqrtx-1)(sqrtx-2)=0`
`<=> [(2sqrtx-1=0),(sqrtx-2=0):} <=>` \(\left[ \begin{array}{l}\sqrt{x}=\dfrac{1}{2}\\\sqrt{x}=2\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{4}\\x=4\end{array} \qquad \text{(tmđk)} \right.\)
Vậy với `x in {1/4;4}` thì `P=8/9`
