Question

4y^2-5xy+x^2=0 решите пожалуста

Answer 1

Это варажение можно лишь разложить на множители, если нужен и x и y пишите в лс.
4y^2-5xy+x^2=(y-x)*(4*y-x)

Answer 2

Solution to 4y^2-5xy+x^2=0

To solve the equation 4y^2-5xy+x^2=0, we can use the given equation to find the values of x and y. Let's proceed with the solution.

First, let's rewrite the equation in the form of a quadratic equation:

x^2 - 5xy + 4y^2 = 0

Now, we can use the quadratic formula to find the values of x and y. The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

y = (-b ± √(b^2 - 4ac)) / (2a)

Where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

Using the coefficients from the given equation, we can substitute them into the quadratic formula to find the values of x and y.

x = (-(-5) ± √((-5)^2 - 4*1*4)) / (2*1)

y = (-(-5) ± √((-5)^2 - 4*1*4)) / (2*1)

Solving for x and y will give us the solutions to the equation.

x = (5 ± √(25 - 16)) / 2

y = (5 ± √(25 - 16)) / 2

By solving the above equations, we can find the values of x and y.

(#1)