Solve the inequality \(\dfrac{3}{x-2}+\dfrac{4}{x}\ge 1\). Enter the sum of all its integer solutions.
Solution
Domain: \(x\ne0\), \(x\ne2\). Combining the terms gives \(-\dfrac{(x-1)(x-8)}{x(x-2)}\ge0\), or \(\dfrac{(x-1)(x-8)}{x(x-2)}\le0\). The interval method gives \(x\in(0;1]\cup(2;8]\). The integer solutions are \(1,3,4,5,6,7,8\), whose sum is \(34\).
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2
Task
One correct answer
Solve the inequality \(\dfrac{1}{x-5}\lt0\).
Solution
The numerator is positive, so the fraction is negative exactly when \(x-5\lt0\). Hence \(x\lt5\), so \(x\in(-\infty;5)\). Answer: A.
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3
Task
One correct answer
Solve the inequality \(\dfrac{3x}{x+1}\lt\dfrac{7}{x+1}\).
Solution
Move everything to one side: \(\dfrac{3x-7}{x+1}\lt0\). The critical points are \(x=-1\) and \(x=\dfrac73\). By the interval method, the fraction is negative on \((-1;\dfrac73)\). Answer: A.
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4
Task
Numeric answer
Solve the inequality \(\dfrac{x^2+11x+30}{x^2+3x-10}\lt0\). Enter the smallest integer satisfying it. If no such integer exists, enter \(100\).
Solution
Factor: \(x^2+11x+30=(x+5)(x+6)\), \(x^2+3x-10=(x+5)(x-2)\). Domain: \(x\ne-5\), \(x\ne2\). This reduces to \(\dfrac{x+6}{x-2}\lt0\), with \(-5\) still excluded. Thus \(x\in(-6;-5)\cup(-5;2)\). The smallest integer solution is \(-4\).
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5
Task
Numeric answer
Find the greatest integer solution of \(\left|\dfrac{3x+1}{x-3}\right|\lt3\).
Solution
Domain: \(x\ne3\). Since both sides are nonnegative, we may square: \((3x+1)^2\lt9(x-3)^2\). This simplifies to \(60x\lt80\), so \(x\lt\dfrac43\). The greatest integer solution is \(1\).
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6
Task
Numeric answer
Find the number of all integer solutions of \(\dfrac{x^2-x-12}{(x+1)^2}\le0\). If there are infinitely many integer solutions, enter \(100\).
Solution
Domain: \(x\ne-1\). The numerator factors as \((x-4)(x+3)\). The denominator is positive for every admissible \(x\), so we need \((x-4)(x+3)\le0\). Hence \(x\in[-3;4]\), excluding \(-1\). There are \(7\) integer solutions.
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7
Task
One correct answer
Solve the inequality \(\dfrac{(x-6)(x+2)^2}{x-3}\le0\).
Solution
Critical points: \(-2\) is a zero of even multiplicity, \(3\) is excluded, and \(6\) is a zero of the numerator. The factor \((x+2)^2\) does not change sign. The solution is \(\{-2\}\cup(3;6]\). Answer: A.
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8
Task
One correct answer
Solve the inequality \(\dfrac{(5-x)^2}{x^2+x-6}\ge0\).
Solution
The denominator is \(x^2+x-6=(x+3)(x-2)\), so \(x\ne-3,2\). The numerator \((5-x)^2\ge0\) and is zero at \(x=5\). The denominator is positive for \(x\lt-3\) or \(x\gt2\). Hence \(x\in(-\infty;-3)\cup(2;+\infty)\). Answer: C.
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9
Task
One correct answer
Choose the number that satisfies \(\dfrac{5}{x-3}\ge1\).
Solution
Move \(1\) to the left: \(\dfrac{5}{x-3}-1=\dfrac{8-x}{x-3}\ge0\). The critical points are \(3\) and \(8\). The solution is \(x\in(3;8]\). Of the listed numbers, only \(4\) belongs to this interval. Answer: E.
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10
Task
One correct answer
Solve the inequality \(\dfrac{x+3}{x-2}\gt0\).
Solution
Critical points: \(x=-3\) is a zero of the numerator and \(x=2\) is excluded. The fraction is positive when numerator and denominator have the same sign: \(x\lt-3\) or \(x\gt2\). Thus \(x\in(-\infty;-3)\cup(2;+\infty)\). Answer: E.
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11
Task
One correct answer
Solve the inequality \(\dfrac{2x-4}{x+1}\lt0\).
Solution
\(\dfrac{2x-4}{x+1}=\dfrac{2(x-2)}{x+1}\). The critical points are \(x=-1\), which is excluded, and \(x=2\), where the numerator is zero. By the interval method, the fraction is negative on \((-1;2)\). Answer: C.
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12
Task
One correct answer
Solve the inequality \(\dfrac{x+1}{x}\le\dfrac43\).
Solution
Move everything to the left: \(\dfrac{x+1}{x}-\dfrac43\le0\). This gives \(\dfrac{3-x}{3x}\le0\), or \(\dfrac{x-3}{x}\ge0\), with \(x\ne0\). Hence \(x\in(-\infty;0)\cup[3;+\infty)\). Answer: A.
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13
Task
One correct answer
Solve the inequality \(\dfrac{x^2+64}{x-5}\gt0\).
Solution
For every real \(x\), \(x^2+64\gt0\). Therefore the sign of the fraction is determined by the denominator. We need \(x-5\gt0\), so \(x\gt5\). Hence \(x\in(5;+\infty)\). Answer: D.
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14
Task
One correct answer
Solve the inequality \(x+\dfrac{1}{x-3}\gt\dfrac{1}{x-3}-2\).
Solution
Domain: \(x\ne3\). The identical fractions on both sides cancel, leaving \(x\gt-2\). Taking the domain into account gives \(x\in(-2;3)\cup(3;+\infty)\). Answer: E.
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15
Task
Numeric answer
Enter the smallest integer satisfying \(\dfrac{(x-3)(x+10)(x^2+8x-9)}{x^2+8x-9}\lt0\).
Solution
\(x^2+8x-9=(x-1)(x+9)\), so \(x\ne1\) and \(x\ne-9\). On the domain, the common factor cancels and we get \((x-3)(x+10)\lt0\), hence \(-10\lt x\lt3\). Exclude \(-9\) and \(1\). The smallest integer solution is \(-8\).
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16
Task
One correct answer
Determine the number of integer solutions of \(\dfrac{6-x}{3x-9}\ge0\).
Solution
Domain: \(x\ne3\). The critical points are \(x=3\) and \(x=6\). The interval method gives \(x\in(3;6]\). The integer solutions are \(4,5,6\), so there are \(3\). Answer: C.
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17
Task
One correct answer
Solve the inequality \(\dfrac{5x+7}{x-1}\le2\).
Solution
Domain: \(x\ne1\). Move \(2\) to the left: \(\dfrac{5x+7-2(x-1)}{x-1}\le0\), so \(\dfrac{x+3}{x-1}\le0\). The interval method gives \(x\in[-3;1)\). Answer: D.
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18
Task
Numeric answer
Solve the inequality \(\dfrac{x^2-5x+4}{-x^2+6x+7}\gt0\). Enter the number of its integer solutions.
Solution
Factor: \(x^2-5x+4=(x-1)(x-4)\), \(-x^2+6x+7=-(x-7)(x+1)\). The critical points are \(-1,1,4,7\). The interval method gives \(x\in(-1;1)\cup(4;7)\). The integer solutions are \(0,5,6\), so there are \(3\).
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19
Task
One correct answer
Solve the inequality \(\dfrac{x}{2x+3}\gt\dfrac1x\).
Solution
Domain: \(x\ne0\) and \(x\ne-1.5\). Combining the terms gives \(\dfrac{(x+1)(x-3)}{x(2x+3)}\gt0\). The critical points are \(-1.5,-1,0,3\). The interval method gives \((-\infty;-1.5)\cup(-1;0)\cup(3;+\infty)\). Answer: E.
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20
Task
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Solve the inequality \(\dfrac{x^2-5x+6}{|x|+7}\lt0\).
Solution
Since \(|x|+7\gt0\) for every \(x\), the sign of the fraction is determined by the numerator. We have \(x^2-5x+6=(x-2)(x-3)\lt0\). The quadratic is negative between its roots, so \(x\in(2;3)\). Answer: B.
