Đáp án:
`3)2/(5-sqrtx)>0(x>=0,x ne 25)`
Mà `2>0`
`=>5-sqrtx>0`
`<=>sqrtx<5`
`=>x<25`
`=>0<=x<25`
`4)(sqrtx+2)/(sqrtx-1)<0(x>=0,x ne 1)`
Mà `sqrtx+2>=2>0`
`=>sqrtx-1<0`
`<=>sqrtx<1`
`=>x<1`
`=>0<=x<1`
`5)(2sqrtx-5)/(3+sqrtx)<0(x>=0)`
Mà `3+sqrtx>=3>0`
`<=>2sqrtx-5<0`
`<=>2sqrtx-5<0`
`<=>2sqrtx<5`
`<=>sqrtx<5/2`
`<=>x<25/4`
`=>0<=x<25/4`
`6)(2-sqrtx)/(sqrtx-1)>0(x>=0,x ne 1)`
`<=>(sqrtx-2)/(sqrtx-1)<0`
`<=>` \(\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-2<0\\\end{cases}\)
`<=>` \(\begin{cases}\sqrt{x}>1\\\sqrt{x}<2\\\end{cases}\)
`<=>` \(\begin{cases}x>1\\x<4\\\end{cases}\)
`<=>1<x<4`
`7)(sqrtx-2)(3sqrtx+1)<0(x>=0)`
Mà `3sqrtx+1>=1>0`
`<=>sqrtx-2<0`
`<=>sqrtx<2`
`<=>x<4`
`<=>0<=x<4`
`8)x-5sqrtx+6>0`
`<=>(sqrtx-2)(sqrtx-3)>0`
`<=>` \(\left[ \begin{array}{l}\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3>0\\\end{cases}\\\begin{cases}\sqrt{x}-2<0\\\sqrt{x}-3<0\\\end{cases}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\begin{cases}\sqrt{x}>2\\\sqrt{x}>3\\\end{cases}\\\begin{cases}\sqrt{x}<2\\\sqrt{x}<3\\\end{cases}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\begin{cases}x>4\\x>9\\\end{cases}\\\begin{cases}x<4\\x<9\\\end{cases}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x>9\\x<4\end{array} \right.\)
`9)x-7sqrtx+12<0(x>=0)`
`<=>(sqrtx-3)(sqrtx-4)<0`
`<=>` \(\begin{cases}\sqrt{x}-3>0\\\sqrt{x}-4<0\\\end{cases}\)
`<=>` \(\begin{cases}\sqrt{x}>3\\\sqrt{x}<4\\\end{cases}\)
`<=>` \(\begin{cases}x>9\\x<16\\\end{cases}\)
`<=>9<x<16`
