Solve for x over the real numbers:
x^2+4 = 6-x
Subtract 6-x from both sides:
x^2+x-2 = 0
The left hand side factors into a product with two terms:
(x-1) (x+2) = 0
Split into two equations:
x-1 = 0 or x+2 = 0
Add 1 to both sides:
x = 1 or x+2 = 0
Subtract 2 from both sides:
Answer: |
| x = 1 or x = -2
Удачи!
(3x-3)(x+2)<0
(3x-3)(x+2)=0 (приравниваем к нулю)
x1=1 x2= -2
x∈(-2;1)
(5x)⁷ *(5x)⁴ *25/((25x²)⁴ *(125x²)=100
(5x)⁷⁺⁴ *25/((5² *x²)⁴ *5*(25x²))=100
(5x)¹¹ *25/((5x)²)⁴ *5*(5x)²)=100
(5x)¹¹ *25/((5x)²*⁴⁺² *5)=100
(5x)¹¹⁻¹⁰ *(25/5)=100
(5x)*5=100, x=4