Question

(3x-7)^2 больше или равно (7x-3)^2

Answer 1

Ответ: x ≤ -1

Квадраты можно убрать:

3x - 7 ≥ 7x - 3

Переносим иксы в левую сторону, числа - в правую:

-4x ≥ 4

При делении на отрицательное число знак меняется на противоположный (≥ на ≤):

x ≤ -1

Answer 2

To compare the expressions $(3x-7)^2$ and $(7x-3)^2$, we'll need to simplify both of them and then analyze their relationship.

Step 1: Expand both expressions:

$(3x-7)^2 = (3x-7)(3x-7) = 9x^2 - 42x + 49$

$(7x-3)^2 = (7x-3)(7x-3) = 49x^2 - 42x + 9$

Step 2: Now we need to determine when $(3x-7)^2$ is greater than or equal to $(7x-3)^2$:

$9x^2 - 42x + 49 \geq 49x^2 - 42x + 9$

Step 3: To solve this inequality, let's bring all terms to one side of the inequality:

$9x^2 - 42x + 49 - (49x^2 - 42x + 9) \geq 0$

Step 4: Simplify and combine like terms:

$9x^2 - 42x + 49 - 49x^2 + 42x - 9 \geq 0$

$-40x^2 + 40 \geq 0$

Step 5: Factor out common terms:

$40(-x^2 + 1) \geq 0$

Step 6: Now we can find the critical points (where the expression equals zero) by setting $-x^2 + 1$ equal to zero and solving for $x$:

$-x^2 + 1 = 0$

$x^2 = 1$

$x = \pm 1$

Step 7: Plot these critical points on a number line and test intervals:

markdownCopy code
--------------o-----------o--------------
       -1           1

Step 8: Test a value in each interval:

• Test $x = 0$: $-0^2 + 1 = 1 \geq 0$ (True)
• Test $x = -2$: $-(-2)^2 + 1 = -3 \nless 0$ (False)
• Test $x = 2$: $-2^2 + 1 = -3 \nless 0$ (False)

Step 9: Analyze the results:

The inequality $9x^2 - 42x + 49 \geq 49x^2 - 42x + 9$ holds true for $x \leq -1$ and $x \geq 1$.

So, the solution to the inequality is $x \leq -1$ or $x \geq 1$.