Question

Lgx+lg(x+1)=lg(2x^2-6)

Answer 1

lgx+lg(x-1)=lg2

D(y):

x>0, x≠1

lg(x(x-1))=lg2

x(x-1)=2

x²-x-2=0

a-b+c=0 (1+1-2=0) → x=-1, x=2

Ответ: x=2

log₅(x+1)=log₅(4x-5)

D(y): x+1>0, 4x-5>0

x>-1, 4x>5

x>-1, x>1,25

x>1,25

x+1=4x-5

x-4x=-5-1

-3x=-6

x=2

log₂(4-x)=log₂(1-2x)

4-x=1-2x

-x+2x=1-4

x=-3

Answer 2

Ответ: 3

Пошаговое объяснение:

Сумму заменяем произведением,   одз: x>0,   2x^2-6>0

Lg(x^2+x)=Lg(2x^2-6),   x^2+x=2x^2-6,   x^2-x-6=0,  корни  3 и  -2(не подходит по одз),   ответ х=3

Answer 3

To solve the equation lg(x) + lg(x+1) = lg(2x^2 - 6), we need to apply logarithm rules to simplify it first. Then we can solve for x.

Step 1: Combine logarithms using the product rule of logarithms. lg(x) + lg(x+1) = lg(2x^2 - 6) lg(x(x+1)) = lg(2x^2 - 6)

Step 2: Eliminate the logarithm by converting both sides to exponential form. x(x+1) = 2x^2 - 6

Step 3: Expand the right side of the equation. x^2 + x = 2x^2 - 6

Step 4: Move all terms to one side of the equation to set it to zero. x^2 + x - 2x^2 + 6 = 0

Step 5: Combine like terms. -x^2 + x + 6 = 0

Step 6: To solve the quadratic equation, either factor it or use the quadratic formula. Factoring: -(x^2 - x - 6) = 0 -(x - 3)(x + 2) = 0

Setting each factor to zero and solving for x: x - 3 = 0 or x + 2 = 0 x = 3 or x = -2

So, the two solutions for the equation are x = 3 and x = -2.