Question

(x - 2)²-4 (x-2) - 5=0​

Answer 1

Ответ: 1; 7

Объяснение:

$t = x-2;\\t^2-4t-5=0;\\D=16+20=36;\\t_1=\frac{4-\sqrt{36}}{2}=-\frac{2}{2}=-1;\\t_2=\frac{4+\sqrt{36}}{2}=\frac{10}{2}=5;\\\\x_1=t_1+2=-1+2=1;\\x_2=t_2+2=5+2=7;$

Answer 2

Ответ:

x={1;7}  т.е. х₁=1 и  х₂=7

Объяснение:

1-й способ :

(х-2)² - 4(х-2) - 5 = 0

(x-2)² +(x-2) -5(x-2) -5 =0

(x-2) × (x-2+1) - 5(x-2+1) =0

(x-2+1) × (x-2-5) =0

(x-1) × (x-7) =0

Произведение равно нулю, если один из множителей равен нулю. Значит

х-1=0       или   х-7=0

х₁=1                    х₂=7

x={1;7}

2-й способ :

(х-2)² - 4(х-2) - 5 = 0

х-2 = q

q²- 4q - 5 = 0

D = b² - 4ac

a = 1     b = -4    c = -5

D = (-4)² - 4×1×(-5) = 16 - (-20) = 16 + 20 = 36 = 6²

q₁ = (-b - √D) /2a  = ((-(-4)) - √36) / (2×1) = (4-6) / 2 = (-2)/2 = -1

q₂ = (-b + √D) /2a  = ((-(-4)) + √36) / (2×1) = (4+6) / 2 = 10/2 = 5

x₁ = q₁ + 2 = (-1) + 2 = 1

x₂ = q₂ + 2 = 5 + 2 = 7

х₁=1       х₂=7

x={1;7}

Answer 3

To solve the quadratic equation (x - 2)² - 4(x - 2) - 5 = 0, we can start by simplifying the equation:

(x - 2)² - 4(x - 2) - 5 = 0

Expanding the square term (x - 2)² gives us:

(x - 2)(x - 2) - 4(x - 2) - 5 = 0

(x - 2)(x - 2) can be rewritten as (x - 2) * (x - 2), which is equal to (x - 2)²:

(x - 2)² - 4(x - 2) - 5 = 0

(x - 2)(x - 2) - 4(x - 2) - 5 = 0

Now, we can distribute and simplify:

(x² - 4x + 4) - 4x + 8 - 5 = 0

x² - 4x + 4 - 4x + 3 = 0

x² - 8x + 7 = 0

Now, we have a quadratic equation in the form ax² + bx + c = 0, where a = 1, b = -8, and c = 7. To solve this equation, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Plugging in the values, we get:

x = (-(-8) ± √((-8)² - 4 * 1 * 7)) / (2 * 1)

x = (8 ± √(64 - 28)) / 2

x = (8 ± √36) / 2

x = (8 ± 6) / 2

Now, we have two possible solutions:

x₁ = (8 + 6) / 2 = 14 / 2 = 7

x₂ = (8 - 6) / 2 = 2 / 2 = 1

Therefore, the solutions to the quadratic equation (x - 2)² - 4(x - 2) - 5 = 0 are x = 7 and x = 1.