3-2y/2-y-3y-2/2+y+16y-y^2-17/4-y^2=0 - вопрос №297943132232216217420 от УРВ05052001 10.11.2022 17:56

Question

Алгебра: 3-2y/2-y-3y-2/2+y+16y-y^2-17/4-y^2=0

УРВ05052001 · Answer

3-2y/2-y - 3y-2/2+y + 16y-y²-17/4-y² = 0
(3-2y)(2+y)-(3y-2)(2-y)/4-y² + 16y-y²-17/4-y² = 0
6+3y-4y-2y²-6y+3y²+4-2y+16y-y²-17/4-y² = 0
7y-7/4-y² = 0

2x-3/3 - 5x+1/4 = 4-x/2
8x-12-15x-3/12 - 4-x/2 = 0
-7x-5-24-6x/12 = 0
-13x-29/12