<p>Solve the inequality $\frac{2x-{{x}^{2}}-5}{(x+1)(x-3)}\le 0$. Hence, solve the following inequalities:</p>
<ol class="i-list"><li><div class="list-item-paragraph">$\frac{2x-{{(x+2)}^{2}}-1}{(x+3)(x-1)}\le 0$ </div></li><li><div class="list-item-paragraph">$\frac{2\left| x \right|-{{x}^{2}}-5}{(\left| x \right|+1)(\left| x \right|-3)}\le 0$</div></li><li><div class="list-item-paragraph">$\frac{2\ln x-{{(\ln x)}^{2}}-5}{(\ln x+1)(\ln x-3)}\le 0$</div></li><li><div class="list-item-paragraph">$\frac{2{{e}^{x}}-{{e}^{2x}}-5}{({{e}^{x}}+1)({{e}^{x}}-3)}\le 0$</div></li></ol>

Solve the inequality $\frac{2x-{{x}^{2}}-5}{(x+1)(x-3)}\le 0$. Hence, solve the following inequalities:

$\frac{2x-{{(x+2)}^{2}}-1}{(x+3)(x-1)}\le 0$ 

$\frac{2\left| x \right|-{{x}^{2}}-5}{(\left| x \right|+1)(\left| x \right|-3)}\le 0$

$\frac{2\ln x-{{(\ln x)}^{2}}-5}{(\ln x+1)(\ln x-3)}\le 0$

$\frac{2{{e}^{x}}-{{e}^{2x}}-5}{({{e}^{x}}+1)({{e}^{x}}-3)}\le 0$

Inequalities

Question 4cf53e63

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