Move everything to one side: \((x+4)(x-7)-3(x-7)\gt 0\). Factor: \((x-7)(x+1)\gt 0\). The product is positive outside the roots \(-1\) and \(7\), so \(x\in(-\infty;-1)\cup(7;+\infty)\). Answer: E.
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2
Task
One correct answer
Solve the inequality \(10-3x\gt 4\).
Solution
\(10-3x\gt 4\Rightarrow -3x\gt -6\). Dividing by the negative number \(-3\) reverses the inequality: \(x\lt 2\). Thus \(x\in(-\infty;2)\). Answer: E.
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3
Task
One correct answer
Find all values of \(x\) for which \(2-5x\) belongs to the interval \((-3,6)\).
Solution
The condition is equivalent to \(-3\lt 2-5x\lt 6\). From \(-3\lt 2-5x\) we get \(x\lt 1\), and from \(2-5x\lt 6\) we get \(x\gt -0.8\). Hence \(-0.8\lt x\lt 1\). Answer: B.
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4
Task
Numeric answer
Solve the inequality \((18+2x)^2\cdot(x^2+8x+15)\le 0\). Enter the sum of all integer solutions.
Solution
Factor: \((18+2x)^2=4(x+9)^2\) and \(x^2+8x+15=(x+3)(x+5)\). The squared factor is nonnegative and is zero at \(x=-9\). Otherwise we need \((x+3)(x+5)\le0\), so \(x\in[-5,-3]\). The integer solutions are \(-9,-5,-4,-3\), whose sum is \(-21\).
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5
Task
One correct answer
Solve the inequality \(2x\ge x^2\).
Solution
\(2x\ge x^2\iff x^2-2x\le0\iff x(x-2)\le0\). The product is nonpositive between the roots, including the endpoints, so \(x\in[0,2]\). Answer: B.
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6
Task
One correct answer
Solve the inequality \(-\dfrac{x}{5}\gt 5\).
Solution
Multiply by \(5\): \(-x\gt 25\). Multiplying by \(-1\) reverses the inequality: \(x\lt -25\). Thus \(x\in(-\infty,-25)\). Answer: A.
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7
Task
One correct answer
Solve the inequality \(\dfrac1x\le\dfrac13\).
Solution
Move everything to one side: \(\dfrac1x-\dfrac13=\dfrac{3-x}{3x}\le0\), with \(x\ne0\). The critical points are \(0\) and \(3\). By the interval method, \(x\in(-\infty,0)\cup[3,+\infty)\). Answer: D.
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8
Task
One correct answer
It is known that \(a\lt b\). Choose the correct inequality.
Solution
From \(a\lt b\), multiplying by \(-1\) reverses the sign: \(-a\gt -b\). Adding \(0.5\) gives \(0.5-a\gt 0.5-b\). Answer: E.
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9
Task
One correct answer
Solve the inequality \((x+4)^2\le16\).
Solution
\((x+4)^2\le16\iff |x+4|\le4\). Hence \(-4\le x+4\le4\), so \(-8\le x\le0\). Thus \(x\in[-8,0]\). Answer: E.
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10
Task
One correct answer
Solve the inequality \(x^3\ge x^2\).
Solution
\(x^3\ge x^2\iff x^2(x-1)\ge0\). Since \(x^2\ge0\), the inequality holds for \(x\ge1\), and also for \(x=0\), where the product is zero. Thus \(x\in\{0\}\cup[1,+\infty)\). Answer: D.
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11
Task
One correct answer
Solve the inequality \(0.2x-54\lt0\).
Solution
\(0.2x-54\lt0\Rightarrow 0.2x\lt54\Rightarrow x\lt270\). Thus \(x\in(-\infty,270)\). Answer: D.
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12
Task
One correct answer
Solve the inequality \((x+4)(x-8)\gt3(x-8)\).
Solution
Move everything to one side: \((x+4)(x-8)-3(x-8)\gt0\). Factoring gives \((x-8)(x+1)\gt0\). The product is positive outside the roots \(-1\) and \(8\), so \(x\in(-\infty,-1)\cup(8,+\infty)\). Answer: A.
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13
Task
One correct answer
Solve the inequality \((x^2+64)(x-5)\gt0\).
Solution
For every real \(x\), \(x^2+64\gt0\), so the sign of the product is determined by \(x-5\). We need \(x-5\gt0\), hence \(x\gt5\). Thus \(x\in(5,+\infty)\). Answer: A.
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14
Task
One correct answer
Solve the inequality \(|x+4|\cdot(x-1)\lt0\).
Solution
We have \(|x+4|\ge0\), and it is zero only at \(x=-4\). For the product to be negative, we need \(x-1\lt0\), so \(x\lt1\), while \(x=-4\) must be excluded because the product is zero there. Thus \(x\in(-\infty,-4)\cup(-4,1)\). Answer: E.
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15
Task
One correct answer
Solve the inequality \(x^3-2x\lt(x+2)(x^2-2x+4)\).
Solution
Using the sum-of-cubes identity, \((x+2)(x^2-2x+4)=x^3+8\). Hence \(x^3-2x\lt x^3+8\Rightarrow -2x\lt8\Rightarrow x\gt-4\). Thus \(x\in(-4,+\infty)\). Answer: A.
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16
Task
One correct answer
Which of the given numbers is a solution of the inequality \(|x|\gt3\)?
Solution
The inequality \(|x|\gt3\) is equivalent to \(x\lt-3\) or \(x\gt3\). Among the given numbers, only \(-8\) satisfies it. Answer: E.
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17
Task
One correct answer
Choose the number that satisfies the inequality \(x^2\lt9\).
Solution
\(x^2\lt9\iff |x|\lt3\iff -3\lt x\lt3\). Of the given numbers, only \(-2\) lies in this interval. Answer: C.
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18
Task
Numeric answer
Solve the inequality \(|x-9|\le3\). Enter the sum of all its integer solutions in the interval \([-15,15]\).
Solution
\(|x-9|\le3\iff -3\le x-9\le3\iff 6\le x\le12\). The integer solutions in the given interval are \(6,7,8,9,10,11,12\). Their sum is \(63\).
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19
Task
One correct answer
Solve the inequality \(4(x-2)-3(x+1)\le2x-2\).
Solution
Expand: \(4x-8-3x-3\le2x-2\), so \(x-11\le2x-2\). Hence \(-9\le x\), i.e. \(x\ge-9\). Thus \(x\in[-9,+\infty)\). Answer: D.
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20
Task
One correct answer
Find the sum of the greatest and the least natural-number solutions of the inequality \(2\le\dfrac{x-1}{2}\le3\).
Solution
Multiply the compound inequality by \(2\): \(4\le x-1\le6\). Adding \(1\) gives \(5\le x\le7\). The natural-number solutions are \(5,6,7\). The least is \(5\), the greatest is \(7\), and their sum is \(12\). Answer: A.
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21
Task
One correct answer
Solve the inequality \( (x-1)(x-3)\le 27-2x \).
Solution
Expand: \(x^2-4x+3\le27-2x\), so \(x^2-2x-24\le0\). Thus \( (x-6)(x+4)\le0\), which is nonpositive between the roots, inclusive. Hence \(x\in[-4,6]\). Answer: A.
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22
Task
One correct answer
Solve the inequality \(4x^2+4x+1\gt0\).
Solution
\(4x^2+4x+1=(2x+1)^2\). A square is positive for every \(x\) except where it is zero: \(2x+1=0\Rightarrow x=-0.5\). Therefore \(x\in(-\infty,-0.5)\cup(-0.5,+\infty)\). Answer: D.
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23
Task
Numeric answer
Solve the inequality \( (x^2-6x+8)(3-x)(x+1)\gt0\). Enter the sum of its integer solutions.
Solution
Factor \(x^2-6x+8=(x-2)(x-4)\) and note that \(3-x=-(x-3)\). This gives \( (x-2)(x-4)(x-3)(x+1)\lt0\). The critical points are \(-1,2,3,4\). By the interval method, \(x\in(-1,2)\cup(3,4)\). The integer solutions are \(0\) and \(1\), whose sum is \(1\).
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24
Task
Numeric answer
Solve the inequality \( (x^3-1)(x^2+2x)(6-x)(49-x^2)\ge0\). Enter the greatest negative integer solution.
Solution
Factor: \(x^3-1=(x-1)(x^2+x+1)\), \(x^2+2x=x(x+2)\), and \(49-x^2=(7-x)(7+x)\). Since \(x^2+x+1\gt0\) for every real \(x\), the inequality is equivalent to \(x(x-1)(x+2)(x-6)(x-7)(x+7)\ge0\). The interval method gives \(x\in(-\infty,-7]\cup[-2,0]\cup[1,6]\cup[7,+\infty)\). The greatest negative integer solution is \(-1\).
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25
Task
One correct answer
Solve the inequality \( (x+1)^7(3-x)^5(x-2)^2\le0\).
Solution
Since \(3-x=-(x-3)\), rewrite the inequality as \( (x+1)^7(x-3)^5(x-2)^2\ge0\). The roots \(-1\) and \(3\) have odd multiplicity, while \(2\) has even multiplicity. The interval method gives \(x\in(-\infty,-1]\cup\{2\}\cup[3,+\infty)\). Answer: C.
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26
Task
One correct answer
Find the smallest solution of \( (x+2)^2(x-1)\ge2(x+2)^2\) on the interval \((-5,5)\).
Solution
Move the right side to the left: \( (x+2)^2[(x-1)-2]\ge0\), so \( (x+2)^2(x-3)\ge0\). At \(x=-2\) the product is zero; for \(x\ne-2\) the squared factor is positive, so we need \(x\ge3\). On \((-5,5)\), the solution set is \(\{-2\}\cup[3,5)\). The smallest solution is \(-2\). Answer: D.
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27
Task
One correct answer
Solve the inequality \(|x+1|\gt1\).
Solution
For \(|u|\gt a\) with \(a\gt0\), we have \(u\lt-a\) or \(u\gt a\). Thus \(x+1\lt-1\) or \(x+1\gt1\), giving \(x\lt-2\) or \(x\gt0\). Hence \(x\in(-\infty,-2)\cup(0,+\infty)\). Answer: C.
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28
Task
One correct answer
Solve the inequality \(|2-x|\lt3\).
Solution
\(|2-x|=|x-2|\), so \(|x-2|\lt3\iff -3\lt x-2\lt3\). Adding \(2\) gives \(-1\lt x\lt5\). Hence \(x\in(-1,5)\). Answer: D.
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29
Task
One correct answer
Solve the inequality \(|x-1|+|x-2|\le3\).
Solution
Split at \(1\) and \(2\). If \(x\le1\), then \(|x-1|+|x-2|=3-2x\le3\Rightarrow x\ge0\), giving \([0,1]\). On \([1,2]\), the sum equals \(1\), so the whole interval works. If \(x\ge2\), then \(|x-1|+|x-2|=2x-3\le3\Rightarrow x\le3\), giving \([2,3]\). Altogether, \(x\in[0,3]\). Answer: A.
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30
Task
One correct answer
Solve the inequality \(|2x|\ge4\).
Solution
\(|2x|=2|x|\), so \(2|x|\ge4\iff |x|\ge2\). Thus \(x\le-2\) or \(x\ge2\). Hence \(x\in(-\infty,-2]\cup[2,+\infty)\). Answer: C.
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31
Task
One correct answer
Among the given numbers, choose the smallest one that satisfies \(2x+7\gt27\).
Solution
\(2x+7\gt27\Rightarrow2x\gt20\Rightarrow x\gt10\). Of the listed numbers, \(15\) and \(20\) work; the smaller is \(15\). Answer: D.
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32
Task
One correct answer
Solve the inequality \(2x-1\ge0\).
Solution
\(2x-1\ge0\Rightarrow2x\ge1\Rightarrow x\ge\dfrac12\). Hence \(x\in[\dfrac12,+\infty)\). Answer: D.
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33
Task
One correct answer
Which of the given numbers is a solution of \(|-2x-3|\gt5\)?
Solution
For \(x=2\), \(|-2\cdot2-3|=|-7|=7\gt5\). Therefore \(2\) is a solution. Answer: E.
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34
Task
One correct answer
Solve the inequality \(x+3\le0\).
Solution
\(x+3\le0\Rightarrow x\le-3\). Hence \(x\in(-\infty,-3]\). Answer: C.
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35
Task
One correct answer
Solve the inequality \(-x^2-x+6\lt0\).
Solution
Multiply by \(-1\), reversing the inequality: \(x^2+x-6\gt0\). Thus \((x+3)(x-2)\gt0\), so the product is positive outside the roots: \(x\in(-\infty,-3)\cup(2,+\infty)\). Answer: C.
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36
Task
One correct answer
Solve the inequality \(3x-7\le8\).
Solution
\(3x-7\le8\Rightarrow3x\le15\Rightarrow x\le5\). Hence \(x\in(-\infty,5]\). Answer: A.
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37
Task
One correct answer
Solve the inequality \((x-5)(x+2)\ge0\).
Solution
The roots are \(x=-2\) and \(x=5\). The product is nonnegative outside the interval between the roots, including the roots. Thus \(x\in(-\infty,-2]\cup[5,+\infty)\). Answer: C.
\(|2x+1|\le5\iff -5\le2x+1\le5\). Subtract \(1\): \(-6\le2x\le4\); divide by \(2\): \(-3\le x\le2\). Hence \(x\in[-3,2]\). Answer: A.
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40
Task
Numeric answer
Solve the inequality \((x+4)(x-1)(x-3)\le0\). Enter the sum of all integer solutions on \([-6,4]\).
Solution
The critical points are \(-4,1,3\). For a cubic polynomial with positive leading coefficient, the signs alternate, so the inequality holds for \(x\in(-\infty,-4]\cup[1,3]\). On \([-6,4]\), the integer solutions are \(-6,-5,-4,1,2,3\), whose sum is \(-9\).
