Question

2) 26 + (2 + x)^3 < x^2 (6 + x);​

Answer 1

Ответ:

Пошаговое объяснение

26+(2+x)^3 < x^2 (6+x)

26+(2+x)(4+4x+x^2) < 6x^2+x^3

26+8+4x+8x+4x^2+2x^2+x^3 < 6x^2+x^3

34+12x+6x^2+x^3 < 6x^2+x^3

Answer 2

To solve the inequality $26 + (2 + x)^3 < x^2 (6 + x)$, we'll first simplify the expressions on both sides and then solve for the values of $x$ that satisfy the inequality.

Step 1: Simplify the inequality. Expand the cube on the left side: $(2 + x)^3 = (2 + x)(2 + x)(2 + x) = (2 + x)(4 + 4x + x^2) = 8 + 12x + 6x^2 + x^3$

The inequality becomes: $26 + 8 + 12x + 6x^2 + x^3 < x^2 (6 + x)$

Step 2: Move all terms to one side of the inequality to set it to zero: $x^3 + 6x^2 + 12x + 34 - x^2 (6 + x) < 0$

Step 3: Simplify further: $x^3 + 6x^2 + 12x + 34 - 6x^2 - x^3 < 0$

Step 4: Cancel out $x^3$ and $6x^2$ terms: $12x + 34 < 0$

Step 5: Move constant term to the other side: $12x < -34$

Step 6: Isolate $x$: $x < \frac{-34}{12}$

Step 7: Reduce the fraction: $x < -\frac{17}{6}$

Thus, the solution to the inequality is $x < -\frac{17}{6}$.