Question

1) x-37x+27=04) x+41x-371=0​

Answer 1

1)-36х=-27
х=27/36 сокращаем и получаем 3/4
2)42х=371
х=371/42

Answer 2

It seems like you've provided two quadratic equations, but there might be some typographical errors in your equations. Let me correct them and solve them for you:

1. x^2 - 37x + 27 = 0

To solve this quadratic equation, you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 1, b = -37, and c = 27. Plugging these values into the formula:

x = (37 ± √((-37)² - 4 * 1 * 27)) / (2 * 1)

x = (37 ± √(1369 - 108)) / 2

x = (37 ± √1261) / 2

So, the solutions for this equation are:

x₁ = (37 + √1261) / 2 x₂ = (37 - √1261) / 2

2. x^2 + 41x - 371 = 0

To solve this quadratic equation, again use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 1, b = 41, and c = -371. Plugging these values into the formula:

x = (-41 ± √(41² - 4 * 1 * (-371))) / (2 * 1)

x = (-41 ± √(1681 + 1484)) / 2

x = (-41 ± √3165) / 2

So, the solutions for this equation are:

x₁ = (-41 + √3165) / 2 x₂ = (-41 - √3165) / 2

These are the solutions to the corrected quadratic equations you provided.