Question

3 x(x+2)-4(4-x)(x+4)> или =5(x²+1)-4(x-+)

Answer 1

Ответ:

${7x}^{2 } + 6x - 64$

Answer 2

To solve the inequality $3x(x+2) - 4(4-x)(x+4) \geq 5(x^2+1) - 4(x-+)$, you'll want to simplify and then solve for (x:

Step 1: Simplify both sides of the inequality.

$$3x(x+2) - 4(4-x)(x+4) \geq 5(x^2+1) - 4(x-+)$$

Step 2: Continue simplifying.

$$3x^2 + 6x - 4(4-x)(x+4) \geq 5x^2 + 5 - 4x-4$$

Step 3: Distribute the terms.

$$3x^2 + 6x - (16 - 4x)(x+4) \geq 5x^2 + 5 - 4x-4$$

Step 4: Expand the expressions.

$$3x^2 + 6x - (16x - 4x^2 + 64) \geq 5x^2 + 5 - 4x-4$$

Step 5: Remove parentheses.

$$3x^2 + 6x - 16x + 4x^2 - 64 \geq 5x^2 + 5 - 4x - 4$$

Step 6: Combine like terms.

$$7x^2 - 22x - 64 \geq 5x^2 + 5 - 4x - 4$$

Step 7: Subtract $5x^2$, $5$, and $4x$ from both sides.

$$2x^2 - 18x - 64 \geq 0$$

Step 8: Divide the entire inequality by 2 to simplify further.

$$x^2 - 9x - 32 \geq 0$$

Step 9: Factor the quadratic on the left side.

$$(x-13)(x+3) \geq 0$$

Step 10: Find the critical points by setting $x-13$ and $x+3$ equal to zero.

$$x - 13 = 0 \implies x = 13$$

$$x + 3 = 0 \implies x = -3$$

Step 11: Create a sign chart and test intervals.

• Choose test values for each interval:

  • Test a value less than -3, e.g., $x = -4$.
  • Test a value between -3 and 13, e.g., $x = 0$.
  • Test a value greater than 13, e.g., $x = 14$.

• Plug the test values into the factored inequality $(x-13)(x+3) \geq 0$ to determine the sign of the expression.

  • For $x = -4$: $((-4)-13)((-4)+3) \geq 0$ $(-17)(-1) \geq 0$ $17 \geq 0$ - True

  • For $x = 0$: $(0-13)(0+3) \geq 0$ $(-13)(3) \geq 0$ $-39 \geq 0$ - False

  • For $x = 14$: $(14-13)(14+3) \geq 0$ $(1)(17) \geq 0$ $17 \geq 0$ - True

Step 12: Analyze the sign chart.

• The solutions to the inequality are $x \leq -3$ and $x \geq 13$.

So, the solution to the inequality $3x(x+2) - 4(4-x)(x+4) \geq 5(x^2+1) - 4(x-+)$ is:

$$x \leq -3 \text{ or } x \geq 13$$