Question

(6x-1)^2-(3x-5)^2=0 Решить уравнение

Answer 1

$(6x-1)^2-(3x-5)^2=0\\\\36x^2-12x+1-(9x^2-30x+25)=0\\\\36x^2-12x+1-9x^2+30x-25=0\\\\27x^2+18x-24=0|:3\\\\9x^2+6x-8=0\\\\D=36-4*9*(-8)=36+288=324\\\\x1=\frac{-6+18}{18} =\frac{12}{18} =\frac{2}{3} \\\\x2=\frac{-6-18}{18} =\frac{-24}{18} =\frac{-4}{3} =-1\frac{1}{3}\\\\Otvet: -1\frac{1}{3}; \frac{2}{3}$

Answer 2

Как-то так. Думаю все правильно

Answer 3

To solve the given equation, we'll use the difference of squares identity, which states that (a^2 - b^2) = (a + b)(a - b). Applying this identity to the equation:

(6x - 1)^2 - (3x - 5)^2 = 0

we can rewrite it as:

[(6x - 1) + (3x - 5)][(6x - 1) - (3x - 5)] = 0

Now, let's simplify further:

(9x - 6)(3x + 4) = 0

Now, set each factor equal to zero and solve for x:

1. 9x - 6 = 0 9x = 6 x = 6/9 x = 2/3

2. 3x + 4 = 0 3x = -4 x = -4/3

So, the solutions to the equation are x = 2/3 and x = -4/3.