88494
The solution set of the in equation \(\frac{x^2+6 x-7}{|x+4|}\lt 0\) is \(\qquad\)
1 \((-7,-4)\)
2 \((-7,-4) \cup(4,1)\)
3 \((-7,1)\)
4 \((-7,-4) \cup(-4,1)\)
Explanation:
(D) : \(\frac{x^{2}+6 x-7}{|x+4|}\lt 0\) \(x^{2}+6 x-7=x^{2}+7 x-x-7\) \(=\mathrm{x}(\mathrm{x}+7)-1(\mathrm{x}+7)\) \(=(\mathrm{x}+7)(\mathrm{x}-1)\) \(\therefore \frac{(\mathrm{x}+7)(\mathrm{x}-1)}{|\mathrm{x}+4|}\lt 0\) case 1: If \(x+4>0,|x+4|=x+4\) \(\frac{(x+7)(x-1)}{x+4}\lt 0\) \(\therefore \mathrm{x} \in(-4,1)\) case 2:- If \(x+4\lt 0\) \(x\lt -4\) and \(|x+4|=-(x+4)\) The given inequality can be written as: \(\frac{(\mathrm{x}+7)(\mathrm{x}-1)}{(\mathrm{x}+4)}>0 ; \quad \mathrm{x}\lt -4\) using method of interval, we get- \(\therefore \mathrm{x} \in(-7,-4)\) The resultant solution is \(x \in(-7,-4) \cup(-4,1)\)
88494
The solution set of the in equation \(\frac{x^2+6 x-7}{|x+4|}\lt 0\) is \(\qquad\)
1 \((-7,-4)\)
2 \((-7,-4) \cup(4,1)\)
3 \((-7,1)\)
4 \((-7,-4) \cup(-4,1)\)
Explanation:
(D) : \(\frac{x^{2}+6 x-7}{|x+4|}\lt 0\) \(x^{2}+6 x-7=x^{2}+7 x-x-7\) \(=\mathrm{x}(\mathrm{x}+7)-1(\mathrm{x}+7)\) \(=(\mathrm{x}+7)(\mathrm{x}-1)\) \(\therefore \frac{(\mathrm{x}+7)(\mathrm{x}-1)}{|\mathrm{x}+4|}\lt 0\) case 1: If \(x+4>0,|x+4|=x+4\) \(\frac{(x+7)(x-1)}{x+4}\lt 0\) \(\therefore \mathrm{x} \in(-4,1)\) case 2:- If \(x+4\lt 0\) \(x\lt -4\) and \(|x+4|=-(x+4)\) The given inequality can be written as: \(\frac{(\mathrm{x}+7)(\mathrm{x}-1)}{(\mathrm{x}+4)}>0 ; \quad \mathrm{x}\lt -4\) using method of interval, we get- \(\therefore \mathrm{x} \in(-7,-4)\) The resultant solution is \(x \in(-7,-4) \cup(-4,1)\)
88494
The solution set of the in equation \(\frac{x^2+6 x-7}{|x+4|}\lt 0\) is \(\qquad\)
1 \((-7,-4)\)
2 \((-7,-4) \cup(4,1)\)
3 \((-7,1)\)
4 \((-7,-4) \cup(-4,1)\)
Explanation:
(D) : \(\frac{x^{2}+6 x-7}{|x+4|}\lt 0\) \(x^{2}+6 x-7=x^{2}+7 x-x-7\) \(=\mathrm{x}(\mathrm{x}+7)-1(\mathrm{x}+7)\) \(=(\mathrm{x}+7)(\mathrm{x}-1)\) \(\therefore \frac{(\mathrm{x}+7)(\mathrm{x}-1)}{|\mathrm{x}+4|}\lt 0\) case 1: If \(x+4>0,|x+4|=x+4\) \(\frac{(x+7)(x-1)}{x+4}\lt 0\) \(\therefore \mathrm{x} \in(-4,1)\) case 2:- If \(x+4\lt 0\) \(x\lt -4\) and \(|x+4|=-(x+4)\) The given inequality can be written as: \(\frac{(\mathrm{x}+7)(\mathrm{x}-1)}{(\mathrm{x}+4)}>0 ; \quad \mathrm{x}\lt -4\) using method of interval, we get- \(\therefore \mathrm{x} \in(-7,-4)\) The resultant solution is \(x \in(-7,-4) \cup(-4,1)\)
88494
The solution set of the in equation \(\frac{x^2+6 x-7}{|x+4|}\lt 0\) is \(\qquad\)
1 \((-7,-4)\)
2 \((-7,-4) \cup(4,1)\)
3 \((-7,1)\)
4 \((-7,-4) \cup(-4,1)\)
Explanation:
(D) : \(\frac{x^{2}+6 x-7}{|x+4|}\lt 0\) \(x^{2}+6 x-7=x^{2}+7 x-x-7\) \(=\mathrm{x}(\mathrm{x}+7)-1(\mathrm{x}+7)\) \(=(\mathrm{x}+7)(\mathrm{x}-1)\) \(\therefore \frac{(\mathrm{x}+7)(\mathrm{x}-1)}{|\mathrm{x}+4|}\lt 0\) case 1: If \(x+4>0,|x+4|=x+4\) \(\frac{(x+7)(x-1)}{x+4}\lt 0\) \(\therefore \mathrm{x} \in(-4,1)\) case 2:- If \(x+4\lt 0\) \(x\lt -4\) and \(|x+4|=-(x+4)\) The given inequality can be written as: \(\frac{(\mathrm{x}+7)(\mathrm{x}-1)}{(\mathrm{x}+4)}>0 ; \quad \mathrm{x}\lt -4\) using method of interval, we get- \(\therefore \mathrm{x} \in(-7,-4)\) The resultant solution is \(x \in(-7,-4) \cup(-4,1)\)
88494
The solution set of the in equation \(\frac{x^2+6 x-7}{|x+4|}\lt 0\) is \(\qquad\)
1 \((-7,-4)\)
2 \((-7,-4) \cup(4,1)\)
3 \((-7,1)\)
4 \((-7,-4) \cup(-4,1)\)
Explanation:
(D) : \(\frac{x^{2}+6 x-7}{|x+4|}\lt 0\) \(x^{2}+6 x-7=x^{2}+7 x-x-7\) \(=\mathrm{x}(\mathrm{x}+7)-1(\mathrm{x}+7)\) \(=(\mathrm{x}+7)(\mathrm{x}-1)\) \(\therefore \frac{(\mathrm{x}+7)(\mathrm{x}-1)}{|\mathrm{x}+4|}\lt 0\) case 1: If \(x+4>0,|x+4|=x+4\) \(\frac{(x+7)(x-1)}{x+4}\lt 0\) \(\therefore \mathrm{x} \in(-4,1)\) case 2:- If \(x+4\lt 0\) \(x\lt -4\) and \(|x+4|=-(x+4)\) The given inequality can be written as: \(\frac{(\mathrm{x}+7)(\mathrm{x}-1)}{(\mathrm{x}+4)}>0 ; \quad \mathrm{x}\lt -4\) using method of interval, we get- \(\therefore \mathrm{x} \in(-7,-4)\) The resultant solution is \(x \in(-7,-4) \cup(-4,1)\)
