Question

Решите уравнение раскрыв модуль по промежуткам |x2 + 2x -3|=| x+ 1| + 2

Answer 1

|x²+2x-3|=|x+1|+2
|(x-1)(x+3)|=|x+1|+2

___________-3____________-1_____________1_____________

1) x≤-3  (x-1)(x+3)=-(x+1)+2
             x²+2x-3=-x-1+2
             x²+2x+x-3-1=0
            x²+3x-4=0
            x(1)=-4∈(-∞;-3]
            x(2)=1∉(-∞;-3]
            x= -4
2) -3<x≤-1  -(x-1)(x+3)=-(x+1)+2
                   -x²-2x+3=-x-1+2
                   -x²-2x+x+3-1=0
                  -x²-x+2=0
                   x²+x-2=0
                   x(1)=-2∈(-3;-1]
                   x(2)=1∉(-3;-1]
                   x=-2
3) -1<x≤1    -(x-1)(x+3)=x+1+2
                   -x²-2x+3-x-3=0
                   -x²-3x=0
                   x²+3x=0
                   x(x+3)=0
                   x(1)=0 ∈(-1;1]
                   x(2)=-3∉(-1;1]
                  x=0
4) x>1        x²+2x-3=x+1+2
                  x²+2x-x-3-3=0
                  x²+x=0
                 x(x+1)=0
                 x(1)=0∉(1;+∞)
                 x(2)=-1∉(1;=∞)
Ответ: -4; -2; 0