# Test: Inequalities II - Ambitious

Double click on maths expressions to zoom
Question 1:   Solve the inequality:  $\frac{{x}^{2}-7x+10}{x-5}\ge 0$
$x\in \left(2;+\infty \right)\\left\{5\right\}$
$x\in \left[2;10\right)\\left\{5\right\}$
$x\in \left[2;+\infty \right)\\left\{5\right\}$
$x\in \left[2;+\infty \right)$
Question 2:   Solve the inequality:  $\frac{x+1}{2x+1}\le 1$
$x\in \left[0;+\infty \right)$
$x\in \left(-\infty ;-\frac{1}{2}\right)\cup \left[0;+\infty \right)$
$x\in \left(-\infty ;-\frac{1}{2}\right)$
$x\in \left(-\infty ;-\frac{1}{2}\right]\cup \left[0;+\infty \right)$
Question 3:   Solve the inequality:  $\frac{\left(x-1\right)\left(x-3\right)}{{x}^{2}+4}<0$
$x\in \left[1;3\right)$
$x\in \left[1;3\right]$
$x\in \left(1;3\right]$
$x\in \left(1;3\right)$
Question 4:   Solve the inequality:  ${x}^{2}-5x<0$
$x\in \left(0;5\right]$
$x\in \left(0;5\right)$
$x\in \left(1;5\right)$
$x\in \left(5;+\infty \right)$
Question 5:   Solve the inequality:  $\frac{{x}^{2}}{2x-5}<1$
$x\in \left(-\infty ;\frac{5}{2}\right)$
$x\in \left[-\frac{5}{2};\frac{5}{2}\right]$
$x\in \left(1;\frac{5}{2}\right)$
$x\in \left(-\infty ;\frac{5}{2}\right]$
Question 6:   Solve the inequality:  $\sqrt{3x-1}<7$
$x\in \left[\frac{1}{3};4\right)$
$x\in \left(\frac{1}{3};16\right]$
$x\in \left[\frac{1}{3};16\right)$
$x\in \left(\frac{1}{3};16\right)$
Question 7:   Solve the inequality:  $\left|x+2\right|\le 1$
$x\in \left[-4;0\right]$
$x\in \left(-3;-1\right)$
$x\in \left[-3;-1\right]$
$x\in \left(-3;-1\right]$
Question 8:   Solve the inequality:  ${x}^{2}-5x+10\le 6$
$x\in \left(-4;+\infty \right)$
$x\in \left[1;4\right]$
$x\in \left(-\infty ;-4\right]\cup \left[1;4\right]$
$x\in \left[1;4\right)$
Question 9:   Solve the inequality:  $\frac{7-2x}{x-3}\le 0$
$x\in \left(-\infty ;3\right)$
$x\in \left(-\infty ;3\right)\cup \left(\frac{7}{2};+\infty \right)$
$x\in \left(-\infty ;3\right)\cup \left[\frac{7}{2};+\infty \right)$
$x\in \left(-\infty ;3\right]\cup \left[\frac{7}{2};+\infty \right)$
Question 10:   Solve the inequality:  $5{x}^{2}-5x\left(x+1\right)\le 45$
$x\in \left[-9;10\right)$
$x\in \left[-9;9\right]$
$x\in \left[-9;+\infty \right)$
$x\in \left[-3;+\infty \right)$