Đáp án+Giải thích các bước giải:
$b/$
Với `x≥0`
Ta có:
`B=1/(\sqrtx+1)-(2-\sqrtx)/(x\sqrtx+1)`
`\to B=(x-\sqrtx+1-(2-\sqrtx))/((\sqrtx+1)(x-\sqrtx+1))`
`\to B=(x-\sqrtx+1-2+\sqrtx)/((\sqrtx+1)(x-\sqrtx+1))`
`\to B=(x-1)/((\sqrtx+1)(x-\sqrtx+1))`
Vậy ` B=(x-1)/((\sqrtx+1)(x-\sqrtx+1))`
$c/$
`P=(A-1).B`
`\to P=((x-1)/(\sqrtx-2)-1).(x-1)/((\sqrtx+1)(x-\sqrtx+1))`
`\to P=(x-1-(\sqrtx-2))/(\sqrtx-2).(x-1)/((\sqrtx+1)(x-\sqrtx+1))`
`\to P=(x-1-\sqrtx+2)/(\sqrtx-2).(x-1)/((\sqrtx+1)(x-\sqrtx+1))`
`\to P=(x-\sqrtx+1)/(\sqrtx-2).(x-1)/((\sqrtx+1)(x-\sqrtx+1))`
`\to P=(x-1)/((\sqrtx-2)(\sqrtx+1)`
Để `P>=2`
`⇔(x-1)/((\sqrtx-2)(\sqrtx+1)>=2`
`⇔(x-1)/((\sqrtx-2)(\sqrtx+1)-2>=0`
`⇔(x-1-2(\sqrtx-2)(\sqrtx+1))/((\sqrtx-2)(\sqrtx+1)>=0`
`⇔(x-1-2(x-\sqrtx-2))/((\sqrtx-2)(\sqrtx+1)≥0`
`⇔(x-1-2x+2\sqrtx+4)/((\sqrtx-2)(\sqrtx+1)≥0`
`⇔(-x+2\sqrtx+3)/((\sqrtx-2)(\sqrtx+1)≥0`
`⇔(-(\sqrtx-3)(\sqrtx+1))/((\sqrtx-2)(\sqrtx+1)≥0`
`⇔(3-\sqrtx)/(\sqrtx-2)≥0`
\(⇔\left[ \begin{array}{l}\begin{cases}3-\sqrt x≥0\\\sqrt x-2>0\end{cases}\\\begin{cases}3-\sqrt x≤0\\\sqrt x-2<0\end{cases}\end{array} \right.\)
\(⇔\left[ \begin{array}{l}\begin{cases}\sqrt x≤3\\\sqrt x>2\end{cases}\\\begin{cases}\sqrt x≥3\\\sqrt x<2\end{cases}\end{array} \right.\)
\(⇔\left[ \begin{array}{l}\begin{cases} x≤9\\ x>4\end{cases}\\\begin{cases} x≥9\\ x<4\end{cases}(loại)\end{array} \right.\)
`⇔4<x≤9` thoả mãn điều kiện `x≥0;x\ne4`
Vậy `4<x≤9` để `P≥2`
