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1) To solve the equation √(2x^2 - x + 3) + 1 = 0, we can start by isolating the square root term:

√(2x^2 - x + 3) = -1

Next, we can square both sides of the equation to eliminate the square root:

2x^2 - x + 3 = (-1)^2

Simplifying, we have:

2x^2 - x + 3 = 1

Rearranging the equation, we get:

2x^2 - x + 2 = 0

This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

x = (-(-1) ± √((-1)^2 - 4(2)(2))) / (2(2))

Simplifying further:

x = (1 ± √(1 - 16)) / 4

Since the discriminant (1 - 16) is negative, the equation has no real solutions. Therefore, there are no solutions to the equation √(2x^2 - x + 3) + 1 = 0.

2) To solve the equation √(2x^2 - x + 3) + 1 = 2x, we can follow a similar process as in the previous equation:

√(2x^2 - x + 3) = 2x - 1

Squaring both sides:

2x^2 - x + 3 = (2x - 1)^2

Expanding the right side:

2x^2 - x + 3 = 4x^2 - 4x + 1

Rearranging the equation:

2x^2 - 3x + 2 = 0

This is a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula. Let's use factoring:

(2x - 1)(x - 2) = 0

Setting each factor equal to zero:

2x - 1 = 0 or x - 2 = 0

Solving for x:

2x = 1 or x = 2

x = 1/2 or x = 2

Therefore, the solutions to the equation √(2x^2 - x + 3) + 1 = 2x are x = 1/2 and x = 2.

3) To solve the equation √(-2x^2 + 6 - x) = 1, we can follow a similar process:

√(-2x^2 + 6 - x) = 1

Squaring both sides:

-2x^2 + 6 - x = 1^2

Simplifying:

-2x^2 + 6 - x = 1

Rearranging the equation:

-2x^2 - x + 5 = 0

This is a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula. Let's use factoring:

(-2x + 5)(x - 1) = 0

Setting each factor equal to zero:

-2x + 5 = 0 or x - 1 = 0

Solving for x:

-2x = -5 or x = 1

x = 5/2 or x = 1

Therefore, the solutions to the equation √(-2x^2 + 6 - x) = 1 are x = 5/2 and x = 1.