1.||x-1|-1|=|x-2|2.|2x-2|-|x|≥x

Предыдущий
вопрос Следующий
вопрос

Case 1: x ≤ 2
In this case, x-1 is negative, so we can rewrite the equation as -(-x-1)-1=|x-2|.
Simplifying this gives us x-1-1=|x-2), or x-2=|x-2|.
Now we can solve this equation:
If x-2 is positive, then x-2=x-2, which gives x=2.
If x-2 is negative, then x-2=-(x-2), which gives x=1.

Case 2: x > 2
In this case, x-1 is positive, so we can rewrite the equation as (-x-1)-1=|x-2|.
Simplifying this gives us -x-1-1=|x-2), or -x-2=|x-2|.
Now we can solve this equation:
If -x-2 is positive, then -x-2=x-2, which gives -2=2x, or x=-1 (not valid since x>2).
If -x-2 is negative, then -x-2=-(x-2), which gives -x-2=-x+2, which has no solution in this case.

Therefore, the solutions to the equation |-x-1|-1|=|x-2| are x=1 and x=2.

Case 1: x ≥ 0
In this case, we have 2x-2-x≥x, which simplifies to x-2≥x and then to -2≥0, which is not true.

Case 2: x < 0
In this case, we have 2x-2+x≥x, which simplifies to x-2≥0.
Solving this inequality gives x≥2.

Therefore, the solution to the inequality 2x-2-|x|≥x is x≥2.