Question

1.x2-2018x+2017=0 2.35x2-59x+24=0

Answer 1

Ответ:

x^2-2018x+2017=0

x^2-x-2017x+2017=0

x×(x-1)-2017(x-1)=0

(x-1)×(x-2017)=0

x-1=0

x-2017=0

x1=1

x2=2017

35x^2-59x+24=0

35x^2-24x-35x+24=0

x(35-24)-(35-24)=0

(35x-24)(x-1)=0

35x-24=0

x-1=0

x1=24/35

x2=1

Answer 2

1)

$x^2-2018x+2017=0$

Вспомним теорему Виета:

$x_1+x_2=-b\\x_1\times x_2=c\\\\x_1+x_2=2018\\x_1\times x_2=2017$

2017 - простое число(то есть оно делится только на 1 и на само себя) ⇒

$1\times2017=2017\\-1\times(-2017)=2017$⇒

Корнями уравнения являются числа 1 и 2017

Ответ: x₁=1, x₂=2017

2)

$35x^2-59x+24=0\\D=3481-3360=121=11^2\\\\x{_1}=\displaystyle{\frac{59+11}{2\times35} }=1\\\\x_2=\frac{59-11}{2\times35} =\frac{24}{35}$

Answer 3

To solve the given quadratic equations, we can use the quadratic formula:

For a quadratic equation of the form ax^2 + bx + c = 0, the quadratic formula is:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's solve the equations one by one.

1. x^2 - 2018x + 2017 = 0

Here, a = 1, b = -2018, and c = 2017.

Using the quadratic formula, we can substitute these values:

x = (-(-2018) ± √((-2018)^2 - 4(1)(2017))) / (2(1))

Simplifying further:

x = (2018 ± √(4072324 - 8068)) / 2

x = (2018 ± √(4064256)) / 2

x = (2018 ± 2016) / 2

For the positive root:

x = (2018 + 2016) / 2 = 4034 / 2 = 2017

For the negative root:

x = (2018 - 2016) / 2 = 2 / 2 = 1

Therefore, the solutions for the first equation are x = 2017 and x = 1.

2. 35x^2 - 59x + 24 = 0

Here, a = 35, b = -59, and c = 24.

Using the quadratic formula:

x = (-(-59) ± √((-59)^2 - 4(35)(24))) / (2(35))

Simplifying further:

x = (59 ± √(3481 - 3360)) / 70

x = (59 ± √(121)) / 70

x = (59 ± 11) / 70

For the positive root:

x = (59 + 11) / 70 = 70 / 70 = 1

For the negative root:

x = (59 - 11) / 70 = 48 / 70 = 24 / 35

Therefore, the solutions for the second equation are x = 1 and x = 24/35.