Question

Решите пожалуйста

Answer 1

((x+3)^2-(3x-1)^2)/5+1=x(2x-3)/2
(x^2+6x+9-9x^2+6x-1)/5+1=(2x^2-3x)/2
2(12x+8-8x^2)+10=10x^2-15x
24x+16-16x^2+10=10x^2-15x
26x^2-15x-24x-26=0
26x^2-39x-26=0
2x^2-3x-2=0
D=9+16=25=5^2
x1=(3+5)/4=2
x2=(3-5)/4=-0,5
Ответ: x1=2; x2=-0,5

Answer 2

Solving Mathematical Equations

To solve the equation -2/9 = 5.6, we can follow these steps:

1. Multiply both sides of the equation by 9 to eliminate the fraction: -2 = 5.6 * 9. 2. Simplify the right side of the equation: -2 = 50.4. 3. Since -2 is not equal to 50.4, the equation has no solution.

Therefore, the equation -2/9 = 5.6 has no solution.

Solving Trigonometric Equations

To solve the equation 3cos^2(x) + 11cos(x) + 6 = 0, we can follow these steps:

1. Let cos(x) = a. 2. The equation becomes 3a^2 + 11a + 6 = 0. 3. Use the quadratic formula to solve for a: a = (-b ± √(b^2 - 4ac)) / (2a). - In this case, a = 3, b = 11, and c = 6. - Substitute the values into the quadratic formula: a = (-11 ± √(11^2 - 4 * 3 * 6)) / (2 * 3). - Simplify the equation: a = (-11 ± √(121 - 72)) / 6. - Further simplify: a = (-11 ± √49) / 6. - The two possible values for a are a = -3 and a = -2/3. 4. Since cos(x) = a, the two possible values for cos(x) are cos(x) = -3 and cos(x) = -2/3. 5. To find the values of x, we can use the inverse cosine function: x = ± arccos(a) + 2πn, where n is an integer. - For cos(x) = -3, x = ± arccos(-3) + 2πn. - For cos(x) = -2/3, x = ± arccos(-2/3) + 2πn.

Therefore, the solutions to the equation 3cos^2(x) + 11cos(x) + 6 = 0 are x = ± arccos(-3) + 2πn and x = ± arccos(-2/3) + 2πn.

Solving Algebraic Equations

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