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If $\frac{x^{2}+2 x+7}{2 x+3} < 6, \mathrm{x} \in \mathrm{R}$, then
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2371 Upvotes
Verified Answer
The correct answer is:
$-1 < x < 11$ or $x < -\frac{3}{2}$
$$
\begin{aligned}
& \frac{x^{2}+2 x+7}{2 x+3} < 6 \\
\Rightarrow & \frac{x^{2}+2 x+7}{2 x+3}-6 < 0 \\
\Rightarrow & \frac{x^{2}-10 x-11}{2 x+3} < 0 \\
\Rightarrow & \frac{(x-11)(x+1)}{2 x+3} < 0 \\
\Rightarrow & \frac{(x-11)(x+1)(2 x+3)}{(2 x+3)^{2}} < 0 \\
\Rightarrow &(x-11)(x+1)(2 x+3) < 0 \\
\Rightarrow & x \in\left(-\infty,-\frac{3}{2}\right) \cup(-1,11)
\end{aligned}
$$
\begin{aligned}
& \frac{x^{2}+2 x+7}{2 x+3} < 6 \\
\Rightarrow & \frac{x^{2}+2 x+7}{2 x+3}-6 < 0 \\
\Rightarrow & \frac{x^{2}-10 x-11}{2 x+3} < 0 \\
\Rightarrow & \frac{(x-11)(x+1)}{2 x+3} < 0 \\
\Rightarrow & \frac{(x-11)(x+1)(2 x+3)}{(2 x+3)^{2}} < 0 \\
\Rightarrow &(x-11)(x+1)(2 x+3) < 0 \\
\Rightarrow & x \in\left(-\infty,-\frac{3}{2}\right) \cup(-1,11)
\end{aligned}
$$
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