Помогите..пожалуйста..

1)xkv + x = 6, xkv +x-6=0,x1=-3,x2=-22)6.(3-x)=4(x+5), 18-6x=4x+20, 10x= -2, x= -1/5 = -0,23)6x = 5.(1+2x), 6x = 5+10x, 4x = -5, x = -5/44)4.x.(x+2) - 3.x.(x-1) = 5. (x-1)(x+2)   4xkv + 8x - 3xkv +3x = 5xkv +5x -10   xkv +11x = 5xkv +5x -10   4xkv-6x-10=0, 2xkv - 3x -5=0Eto kvadratičnoe uravnenie imeet korni: 5/2 i -15)(x+3)(x+1)+x(x-1) = 8, xkv + 4x +3 +xkv -x =8, 2xkv+3x-5=0Eto uravnenie imeet korni -5/2 i 16)xkv -3x = x.(x-3), xkv+3x = x.(x+3), xkv-9 = (x+3)(x-3)(2x-5)(x+3) - (x+2)(x-3) + x-5 = 0Dalše uže, dumaju, ty budeš umet .7)(25-ykv)/y-5 = 0 -eto bozmožno tolko togda, kogda 25 - ykv = 0,.t.e.kogda (5+y)(5-y)=0, Korni: 5 i -5