Question

(3x - 7) ² ≥ (7x - 3)²​

Answer 1

Ответ:

9x^2-42x+49>=49x^2-42x+9

9x^2-49x^2-42x+42x>=9-49

-40x^2>=-40

x^2<=1

-1<=x=<1

x [-1;1]

Answer 2

To solve the inequality $(3x - 7)^2 \geq (7x - 3)^2$, we can simplify both sides of the equation and solve for $x$. Let's start by expanding the squares:

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

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

Now, we can rewrite the inequality as:

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

Let's simplify it further:

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

Combining like terms:

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

Dividing both sides by -40 (note: dividing by a negative number reverses the inequality):

$x^2 - 1 \leq 0$

Now, we can solve for $x$ by factoring the quadratic:

$(x - 1)(x + 1) \leq 0$

The solutions to this inequality are $x \leq -1$ and $x \geq 1$.

Therefore, the solution to the original inequality $(3x - 7)^2 \geq (7x - 3)^2$ is $x \leq -1$ and $x \geq 1$.