Question

Неравенства 1) 4-2x<92) x^2-4x-21>03) 1-16x^2>0

Answer 1

) 4-2x<9 => -2x<9-4 => x>2.5
2) x^2-4x-21>0 => D=(-4)^2-4*(-21)=16+84=100=10; x1 = (4+10)/2=7; x2=(4-10)/2=-3; Ответ: (-3;7)
3) 1-16x^2>0 => - 16x^2>-1 => 4x<1 => x< 0.25

Answer 2

To solve these inequalities, we'll find the values of 'x' that satisfy each one. Remember that when solving inequalities, the goal is to isolate 'x' on one side of the inequality sign.

1. $4 - 2x < 9$

Step 1: Subtract 4 from both sides: $4 - 4 - 2x < 9 - 4$

Simplified inequality: $-2x < 5$

Step 2: Divide both sides by -2 (Note: When dividing by a negative number, the inequality sign flips): $\frac{-2x}{-2} > \frac{5}{-2}$

Simplified inequality: $x > -\frac{5}{2}$

So, the solution for this inequality is $x > -\frac{5}{2}$.

2. $x^2 - 4x - 21 > 0$

To solve this quadratic inequality, we first find the critical points where the inequality may change. The critical points are the values of 'x' where the expression $x^2 - 4x - 21$ equals zero.

Step 1: Factor the quadratic expression: $x^2 - 4x - 21 = (x - 7)(x + 3)$

Step 2: Set each factor equal to zero and solve for 'x': $x - 7 = 0 \Rightarrow x = 7$ $x + 3 = 0 \Rightarrow x = -3$

Now we have three intervals to test: a) $x < -3$ b) $-3 < x < 7$ c) $x > 7$

Step 3: Test each interval with a test point to determine if it satisfies the inequality:

a) Test with $x = -4$: $(-4)^2 - 4(-4) - 21 = 16 + 16 - 21 = 11$, which is greater than zero. So, $x < -3$ is a valid solution.

b) Test with $x = 0$: $0^2 - 4(0) - 21 = -21$, which is less than zero. So, $-3 < x < 7$ is not a valid solution.

c) Test with $x = 8$: $8^2 - 4(8) - 21 = 64 - 32 - 21 = 11$, which is greater than zero. So, $x > 7$ is a valid solution.

Therefore, the solution for the inequality $x^2 - 4x - 21 > 0$ is $x < -3$ or $x > 7$.

3. $1 - 16x^2 > 0$

Step 1: Add $16x^2$ to both sides: $1 + 16x^2 > 0$

Step 2: Subtract 1 from both sides: $16x^2 > -1$

Step 3: Divide both sides by 16 (since $16$ is positive, the inequality sign remains unchanged): $x^2 > -\frac{1}{16}$

Since $x^2$ cannot be negative for any real value of 'x', there are no real solutions for this inequality. The inequality $1 - 16x^2 > 0$ has no solution in the real number system.

In summary:

1. $x > -\frac{5}{2}$
2. $x < -3$ or $x > 7$
3. No solution in the real number system.