Question

(4-7x) (x +2)/(x-3) (x-2)> 0

Answer 1

Ответ данного неравенства на приложенной фотографии.

Answer 2

To solve the inequality $(4-7x)(x+2)/(x-3)(x-2) > 0$, we'll follow these steps:

Step 1: Find the critical points by setting the expression equal to zero and solving for $x$. Step 2: Create intervals using the critical points. Step 3: Test the expression within each interval to determine the sign of the expression. Step 4: Identify the intervals where the expression is greater than zero.

Let's proceed with the steps:

Step 1: Find the critical points (where the expression is equal to zero):

$(4-7x)(x+2)/(x-3)(x-2) = 0$

The expression is equal to zero when either $4-7x = 0$ or $x+2 = 0$ or $(x-3) = 0$ or $(x-2) = 0$.

Solving for $x$, we get:

1. $4-7x = 0$ $-7x = -4$ $x = \frac{4}{7}$

2. $x+2 = 0$ $x = -2$

3. $x-3 = 0$ $x = 3$

4. $x-2 = 0$ $x = 2$

Step 2: Create intervals using the critical points:

We have four critical points: $x = \frac{4}{7}$, $x = -2$, $x = 3$, and $x = 2$.

The intervals are:

$(- \infty, -2)$ $(-2, \frac{4}{7})$ $(\frac{4}{7}, 2)$ $(2, 3)$ $(3, \infty)$

Step 3: Test the expression within each interval to determine the sign:

Pick any value within each interval and evaluate the expression.

For $x = -3$ (in $(- \infty, -2)$): $(4-7(-3))(-3+2)/(-3-3)(-3-2) = (-25)(-1)/(-6)(-5) = \frac{5}{6} > 0$

For $x = 0$ (in $(-2, \frac{4}{7})$): $(4-7(0))(0+2)/(0-3)(0-2) = (4)(2)/(-3)(-2) = \frac{4}{3} > 0$

For $x = 1$ (in $(\frac{4}{7}, 2)$): $(4-7(1))(1+2)/(1-3)(1-2) = (-3)(3)/(-2)(-1) = \frac{9}{2} > 0$

For $x = 2.5$ (in $(2, 3)$): $(4-7(2.5))(2.5+2)/(2.5-3)(2.5-2) = (10.5)(4.5)/(-0.5)(0.5) = -189 < 0$

For $x = 4$ (in $(3, \infty)$): $(4-7(4))(4+2)/(4-3)(4-2) = (-20)(6)/(-1)(2) = 60 > 0$

Step 4: Identify the intervals where the expression is greater than zero:

The expression $(4-7x)(x+2)/(x-3)(x-2)$ is greater than zero in the intervals:

$(- \infty, -2)$ $(\frac{4}{7},2)$ 0 0