Question

(х+2)(х+1)-(х-3)(х+4)=12​

Answer 1

Ответ:

$-1$

Объяснение:

$(x+2)(x+1)-(x-3)(x+4)=12\\x^2+2x+x+2-(x^2-3x+4x-12)=12\\x^2+3x+2-(x^2+x-12)=12\\x^2+3x+2-x^2-x+12=12\\2x=-2\\x=-2:2\\x=-1$

Answer 2

To solve the equation (х+2)(х+1)-(х-3)(х+4)=12, let's expand the expressions and simplify:

Expanding the left side: (х+2)(х+1) = х² + х + 2х + 2 = х² + 3х + 2

(х-3)(х+4) = х² + 4х - 3х - 12 = х² + x - 12

Substituting the expanded expressions into the equation: (х² + 3х + 2) - (х² + x - 12) = 12

Simplifying: х² + 3х + 2 - х² - x + 12 = 12

The х² terms cancel out: 3х - x + 14 = 12

Combining like terms: 2х + 14 = 12

Subtracting 14 from both sides: 2х = 12 - 14 2х = -2

Dividing by 2: х = -2/2 х = -1

Therefore, the solution to the equation (х+2)(х+1)-(х-3)(х+4)=12 is х = -1.