Question

(4x+1)²-(1-3x)(1-3x)(1+3x)=(5x+2)²

Answer 1

(4x+1)²-(1-3x)(1-3x)(1+3x)=(5x+2)²

(4х+1)²-(1-3x)²(1+3х)=
=((4х+1)+(х+1))²

(4х+1)²-(1-3x)²(1+3х)=
=(4х+1)²+2(4х+1)(х+1)+(х+1)²
(8х+2+х+1)(х+1)+(1+3х)(1-3x)²=0

3(3х+1)(х+1)+(1+3х)(1-3x)²=0
(3х+1)(3х+3+1-6х+9х²)=0
(3х+1)(9х²-3х+4)=0

3х+1=0
х= - ⅓

9х²-3х+4=0

D = b² - 4ac =
= (-3)² - 4·9·4 =
= 9 - 144 = -135
поэтому данное квадратное уравнение
не имеет действительных корней

Ответ:
х= - ⅓

Answer 2

Let's simplify the given expression step by step:

(4x+1)² - (1-3x)(1-3x)(1+3x) = (5x+2)²

Expanding the square terms:

(16x² + 8x + 1) - (1-3x)(1-3x)(1+3x) = 25x² + 20x + 4

Next, let's expand the term on the right side of the equation:

16x² + 8x + 1 - (1 - 3x)(1 - 3x)(1 + 3x) = 25x² + 20x + 4

Expanding the cube term:

16x² + 8x + 1 - (1 - 6x + 9x²)(1 + 3x) = 25x² + 20x + 4

Now, let's multiply the terms inside the parentheses:

16x² + 8x + 1 - (1 + 3x - 6x - 18x²) = 25x² + 20x + 4

Simplifying further:

16x² + 8x + 1 - 1 - 3x + 6x + 18x² = 25x² + 20x + 4

Combine like terms:

34x² + 11x = 25x² + 20x + 4

Move all terms to one side of the equation:

34x² + 11x - 25x² - 20x - 4 = 0

Combine like terms again:

(34x² - 25x²) + (11x - 20x) - 4 = 0

9x² - 9x - 4 = 0

Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. However, upon examining the equation, it doesn't seem to have simple integer solutions.

If you have any specific questions or if you want to explore other methods of solving the equation, please let me know!