Question

A) (x + 3)(x-8)<x(x-2); в) (4x - 1)(4x + 1) ≤4+16x(x-1);б) (2x - 3)² > 2x(2x - 1);г) (x-3)(x² + 3x +9) > x³ + 2x - 1.​

Answer 1

Ответ:(x + 3)(x-8)<x(x-2)

x^2-8x+3x-24<x^2-2х

-8+3х-24<-2х

-5х-24<-2х

-5х+2х<24

-3х<24

х>-8

4x - 1)(4x + 1) ≤4+16x(x-1);

16х^2-1 більше-рівне 4+16х^2-16х

-1більше-рівне4+1

16хбільше-рівне5

2x - 3)² > 2x(2x - 1);

4х^2-12x+9>4x^2-2x

-12x+9>-2x

-12x+2x>-9

-10x+2x>-9

-10x>-9

(x-3)(x² + 3x +9) > x³ + 2x - 1

x^3-27>x^3+2x-1

-27>2x-1

-2x>-1+27

-2x>26

x<-13

Объяснение:

Answer 2

Let's solve each inequality step by step:

а) $(x + 3)(x - 8) < x(x - 2)$

First, let's expand both sides of the inequality: $x^2 - 5x - 24 < x^2 - 2x$

Now, let's simplify and solve for $x$: $-5x - 24 < -2x$ $-5x + 2x < 24$ $-3x < 24$ $x > -8$

So the solution for this inequality is $x > -8$.

б) $(4x - 1)(4x + 1) \leq 4 + 16x(x - 1)$

Let's expand both sides: $16x^2 - 1 \leq 4 + 16x^2 - 16x$

Now, let's simplify and solve for $x$: $16x^2 - 1 - 4 - 16x^2 + 16x \leq 0$ $16x \leq 3$ $x \leq \frac{3}{16}$

So the solution for this inequality is $x \leq \frac{3}{16}$.

в) $(2x - 3)^2 > 2x(2x - 1)$

Let's expand both sides: $4x^2 - 12x + 9 > 4x^2 - 2x$

Now, let's simplify and solve for $x$: $-12x + 9 > -2x$ $-12x + 2x > -9$ $-10x > -9$ $x < \frac{9}{10}$

So the solution for this inequality is $x < \frac{9}{10}$.

г) $(x - 3)(x^2 + 3x + 9) > x^3 + 2x - 1$

Let's expand both sides: $x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 > x^3 + 2x - 1$

Now, let's simplify and solve for $x$: $-27 > -1$

This is always true, so the solution for this inequality is all real numbers.

If you have any further questions or need additional assistance, feel free to ask.