Question

(5х-2)^2≤(3х+1)^2+(4х-3)^2

Answer 1

Ответ:на фото

Объяснение:

Answer 2

To solve the inequality $(5x-2)^2 \leq (3x+1)^2 + (4x-3)^2$, we can simplify both sides and solve for $x$. Let's begin:

Expanding both sides of the inequality:

$(5x-2)^2 = 25x^2 - 20x + 4$ $(3x+1)^2 = 9x^2 + 6x + 1$ $(4x-3)^2 = 16x^2 - 24x + 9$

Now we can rewrite the inequality:

$25x^2 - 20x + 4 \leq 9x^2 + 6x + 1 + 16x^2 - 24x + 9$

Simplifying the right side:

$25x^2 - 20x + 4 \leq 25x^2 - 18x + 10$

Moving all terms to the left side:

$0 \leq 25x^2 - 20x + 4 - 25x^2 + 18x - 10$

Simplifying further:

$0 \leq -2x - 6$

To isolate $x$, let's multiply both sides by -1:

$0 \geq 2x + 6$

Subtracting 6 from both sides:

$-6 \geq 2x$

Dividing both sides by 2 (keeping in mind that dividing by a negative number flips the inequality):

$-3 \geq x$

So, the solution to the inequality $(5x-2)^2 \leq (3x+1)^2 + (4x-3)^2$ is $x \leq -3$.