(k-9)x^2-(k+12)x<4-এর সমাধান করুন

$$( k - 9 ) x ^ { 2 } - ( k + 12 ) x < 4$$

x এর জন্য সমাধান করুন

$$\left\{\begin{matrix}x\neq -\frac{2}{3}\text{, }&k=0\\x\neq \frac{2}{7}\text{, }&k=-40\\x\in \mathrm{R}\text{, }&k>-40\text{ and }k<0\\x\in \left(-\frac{\sqrt{k\left(k+40\right)}-k-12}{2\left(k-9\right)},\frac{\sqrt{k\left(k+40\right)}+k+12}{2\left(k-9\right)}\right)\text{, }&k>9\\x>-\frac{4}{21}\text{, }&k=9\\x\in \left(-\infty,\frac{\sqrt{k\left(k+40\right)}+k+12}{2\left(k-9\right)}\right)\cup \left(-\frac{\sqrt{k\left(k+40\right)}-k-12}{2\left(k-9\right)},\infty\right)\text{, }&k<-40\text{ or }\left(k>0\text{ and }k<9\right)\end{matrix}\right.$$

k এর জন্য সমাধান করুন

$$\left\{\begin{matrix}k<\frac{\left(3x+2\right)^{2}}{x\left(x-1\right)}\text{, }&x<0\text{ or }x>1\\k\in \mathrm{R}\text{, }&x=1\text{ or }x=0\\k>\frac{\left(3x+2\right)^{2}}{x\left(x-1\right)}\text{, }&x>0\text{ and }x<1\end{matrix}\right.$$