Question

Решите пж примеры : А)КОРЕНЬ В СТЕПЕНИ 3 ПОД КОРНЕМ 2X+3=-3 Б)КОРЕНЬ ПОДКОРНЕМ 2X+3=-3 в)корень подкорнем 2x+3=3 г)корень подконем 4x+3=корень под корнем x квадрате + x - 1 д)корень подкорнем 2x+3=-3-2x спасибо заранее)

Answer 1

Щас постараюсь сделать.....

Answer 2

Solution for Equation A: √(2x+3)^3 = -3

To solve the equation √(2x+3)^3 = -3, we need to isolate the variable x.

1. Start by cubing both sides of the equation to eliminate the square root: (√(2x+3))^3 = (-3)^3 (2x+3)^3 = -27

2. Take the cube root of both sides to remove the exponent: 2x+3 = -3∛27 2x+3 = -3 * (-3) 2x+3 = 9

3. Subtract 3 from both sides to isolate the variable x: 2x = 9 - 3 2x = 6

4. Divide both sides by 2 to solve for x: x = 6/2 x = 3

Therefore, the solution for Equation A is x = 3.

Solution for Equation B: √(2x+3) = -3

To solve the equation √(2x+3) = -3, we need to isolate the variable x.

1. Start by squaring both sides of the equation to eliminate the square root: (√(2x+3))^2 = (-3)^2 2x+3 = 9

2. Subtract 3 from both sides to isolate the variable x: 2x = 9 - 3 2x = 6

3. Divide both sides by 2 to solve for x: x = 6/2 x = 3

Therefore, the solution for Equation B is x = 3.

Solution for Equation C: √(2x+3) = 3

To solve the equation √(2x+3) = 3, we need to isolate the variable x.

1. Start by squaring both sides of the equation to eliminate the square root: (√(2x+3))^2 = 3^2 2x+3 = 9

2. Subtract 3 from both sides to isolate the variable x: 2x = 9 - 3 2x = 6

3. Divide both sides by 2 to solve for x: x = 6/2 x = 3

Therefore, the solution for Equation C is x = 3.

Solution for Equation D: √(4x+3) = √(x^2 + x - 1)

To solve the equation √(4x+3) = √(x^2 + x - 1), we need to isolate the variable x.

1. Square both sides of the equation to eliminate the square roots: (√(4x+3))^2 = (√(x^2 + x - 1))^2 4x+3 = x^2 + x - 1

2. Rearrange the equation to form a quadratic equation: x^2 + x - 1 - (4x + 3) = 0 x^2 - 3x - 4 = 0

3. Solve the quadratic equation by factoring or using the quadratic formula. In this case, we'll use factoring: (x - 4)(x + 1) = 0

Setting each factor equal to zero: x - 4 = 0 or x + 1 = 0

Solving for x: x = 4 or x = -1

Therefore, the solutions for Equation D are x = 4 and x = -1.

Solution for Equation E: √(2x+3) = -3 - 2x

To solve the equation √(2x+3) = -3 - 2x, we need to isolate the variable x.

1. Square both sides of the equation to eliminate the square root: (√(2x+3))^2 = (-3 - 2x)^2 2x+3 = 9 + 12x + 4x^2

2. Rearrange the equation to form a quadratic equation: 4x^2 + 12x + 9 - 2x - 3 = 0 4x^2 + 10x + 6 = 0

3. Solve the quadratic equation by factoring or using the quadratic formula. In this case, we'll use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

For the equation 4x^2 + 10x + 6 = 0, the coefficients are: a = 4, b = 10, c = 6

Plugging these values into the quadratic formula: x = (-10 ± √(10^2 - 4 * 4 * 6)) / (2 * 4) x = (-10 ± √(100 - 96)) / 8 x = (-10 ± √4) / 8 x = (-10 ± 2) / 8

Simplifying the solutions: x = (-10 + 2) / 8 or x = (-10 - 2) / 8 x = -8 / 8 or x = -12 / 8 x = -1 or x = -3/2

Therefore, the solutions for Equation E are x = -1 and x = -3/2.

I hope this helps! Let me know if you have any further questions.