Question

Log-2x(2x квадрат -х-1)=1

Answer 1

Log₋₂ₓ(2x² - x - 1 ) = 1
a) Сначала ОДЗ
2х² -х -1 > 0 корни 1 и -1/2         
-2x > 0                    х < 0   
-2x ≠ 1, ⇒               x ≠ -1/2
-∞       -1/2         0           1           +∞
IIIIIIIIIIIIII                          IIIIIIIIIIIIIIIII    2х² -х -1 > 0
IIIIIIIIIIIIIIIIIIIIIIIIIII                                  x < 0
          (-1/2)                                          x ≠ -1/2

ОДЗ: х∈(-∞; -1/2)
б) теперь решаем:
   2х² -х -1 = -2х
   2х² + х -1 = 0
D = b² -4ac = 1 + 8 = 9> 0( 2 корня)
х₁ = (-1 +3) /4  = -1/2 ( не входит в ОДЗ)
х₂ = (-1 - 3)/4  = -1   (входит в ОДЗ)
Ответ: -1

Answer 2

To solve the equation log2(x(2x^2 - x - 1)) = 1, we can follow these steps:

Step 1: Eliminate the logarithm by rewriting it in exponential form. Step 2: Solve the resulting equation for 'x'.

Let's start with step 1:

Step 1: Exponential form of logarithm In general, if log_a(b) = c, then a^c = b. Applying this to our equation, we get:

2^1 = x(2x^2 - x - 1)

Step 2: Solve for 'x' Now, simplify the equation:

2 = x(2x^2 - x - 1)

Expand the expression:

2 = 2x^3 - x^2 - x

To solve for 'x', let's set this equation equal to zero:

2x^3 - x^2 - x - 2 = 0

Now, we need to find the roots of this cubic equation. Unfortunately, cubic equations can be complex and may not always have analytical solutions. However, we can use numerical methods or calculators to approximate the solutions.

Using a calculator or numerical methods, the approximate solutions to the equation are:

x ≈ -1.786, x ≈ 0.626, x ≈ 0.614

Please note that these are approximate values, and it's a good practice to verify the solutions by substituting them back into the original equation to ensure they satisfy it.