Question

Решите пажалуйста алшгебра

Answer 1

Что решить то? Ничего же нет.

Answer 2

Algebraic Equations

To solve the algebraic equations you provided, let's break them down one by one.

1. Equation 1: - Equation: ^4 - ^3 - 5^2 - - 6 < 0 - To solve this equation, we need more information. The equation seems to be incomplete or missing some terms. Could you please provide the complete equation?

2. Equation 2: - Equation: x^2 + y^2 = 20 - Equation: 3x + y = 2 - To solve this system of equations, we can use the substitution method or the elimination method. Let's use the substitution method: - From the second equation, we can express y in terms of x: y = 2 - 3x. - Substituting this value of y into the first equation, we get: x^2 + (2 - 3x)^2 = 20. - Simplifying the equation, we have: 10x^2 - 12x - 16 = 0. - Solving this quadratic equation, we find two solutions for x: x1 = -0.8 and x2 = 2. - Substituting these values of x back into the second equation, we can find the corresponding values of y: y1 = 4.4 and y2 = -4. - Therefore, the solutions to the system of equations are: - Solution 1: x1 = -0.8, y1 = 4.4 - Solution 2: x2 = 2, y2 = -4.

3. Equation 3: - Equation: xy - 3y^2 = -24 - Equation: xy + 2y^2 = 21 - To solve this system of equations, we can use the elimination method. Let's subtract the second equation from the first equation: - (xy - 3y^2) - (xy + 2y^2) = (-24) - 21 - Simplifying the equation, we have: -5y^2 = -45 - Solving for y^2, we find: y^2 = 9 - Taking the square root of both sides, we get two solutions for y: y1 = -3 and y2 = 3. - Substituting these values of y back into the second equation, we can find the corresponding values of x: x1 = -9 and x2 = 8 1/3. - Therefore, the solutions to the system of equations are: - Solution 1: x1 = -9, y1 = -3 - Solution 2: x2 = 8 1/3, y2 = 3.

4. Equation 4: - Equation: (60/x) + 1 = 60/(x-2) - To solve this equation, we can simplify it by multiplying both sides by x(x-2) to eliminate the denominators: - 60(x-2) + x(x-2) = 60x - Simplifying the equation, we have: x^2 - 2x - 120 = 0 - Factoring the quadratic equation, we find: (x-12)(x+10) = 0 - Solving for x, we get two solutions: x1 = 12 and x2 = -10. - Therefore, the solutions to the equation are: - Solution 1: x1 = 12 - Solution 2: x2 = -10.

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