Question

1) |х^2 + 2х| - |2-х| = |х^2 - х|

Answer 1

x^2+2x-2+x=x^2-x

x^2-x^2+x+2x+x=2

4x=2

x=0.5;-0.5

Answer 2

To solve the equation |x^2 + 2x| - |2 - x| = |x^2 - x|, you can break it down into different cases based on the sign of the expressions inside the absolute value symbols.

1. Case 1: x^2 + 2x and 2 - x are both non-negative.
2. Case 2: x^2 + 2x is non-negative, and 2 - x is negative.
3. Case 3: x^2 + 2x is negative, and 2 - x is non-negative.
4. Case 4: x^2 + 2x and 2 - x are both negative.

Let's solve each case individually:

Case 1: x^2 + 2x ≥ 0 (1) 2 - x ≥ 0 (2)

From (1), we know that x(x + 2) ≥ 0, which means x ≤ -2 or x ≥ 0. From (2), we know that -x ≥ -2, which implies x ≤ 2.

So, in this case, x ≤ -2 or x ≥ 0, and 2 - x ≥ 0. So, x is in the range [0, ∞).

Case 2: x^2 + 2x ≥ 0 (1) 2 - x < 0 (2)

From (1), x(x + 2) ≥ 0, which means x ≤ -2 or x ≥ 0. From (2), -x < -2, which implies x > 2.

So, in this case, x > 2, and 2 - x < 0.

Case 3: x^2 + 2x < 0 (1) 2 - x ≥ 0 (2)

From (1), x(x + 2) < 0, which means -2 < x < 0. From (2), -x ≥ -2, which implies x ≤ 2.

So, in this case, -2 < x < 0, and 2 - x ≥ 0. Therefore, x is in the range (-2, 2].

Case 4: x^2 + 2x < 0 (1) 2 - x < 0 (2)

From (1), x(x + 2) < 0, which means -2 < x < 0. From (2), -x < -2, which implies x > 2.

So, in this case, there is no valid solution because it leads to contradictory inequalities.

Now, let's put it all together:

In Case 1, x is in the range [0, ∞). In Case 2, x is greater than 2. In Case 3, x is in the range (-2, 2].

Therefore, the solution to the equation |x^2 + 2x| - |2 - x| = |x^2 - x| is the combination of these cases:

x ≤ -2 (no solution in this case) -2 < x < 0 (from Case 3) 0 ≤ x < 2 (from Case 1) x > 2 (from Case 2)

So, the solution is:

-2 < x < 0 or 0 ≤ x < 2 or x > 2