Question

Решите уравнение (x-5)^2+2*(x-5)(x+7)+(x+7)^2=0

Answer 1

$(x-5)^2+2(x-5)(x+7)+(x+7)^2=0\\x-5=a \ \ \ \ \ \ \ \ x+7=b\\a^2+2ab+b^2=0\\(a+b)^2=0\\(x-5+x+7)^2=0\\(2x+2)^2=0\\2x+2=0\\2x=-2\\x=-1$

Answer 2

Ответ:

х=-1

Объяснение:

(х5-5)^2+2(x-5)*(x+7)+(x+7)^2=0

(x-5+x+7)^2=0 (разложить на множители используя $a^{2}$+2ab+$b^{2}$=(a+b)^2

(2x+2)^2=0 (x объединяются; вычисляется сумма)

(2(x+1))^2=0 (выносим за скобки общий множитель)

(2x+2)^2=0

2x+2=0 (результатом возведения в степень может быть 0 только тогда когда основание равно 0)

2x=-2 (переносим 2 сменив знак)

x=-1 (сокращаем)

Answer 3

To solve the given quadratic equation, we can simplify it first and then proceed to find the solutions. Let's expand and simplify the equation:

(x - 5)^2 + 2(x - 5)(x + 7) + (x + 7)^2 = 0

To simplify, let's use the formula (a + b)^2 = a^2 + 2ab + b^2:

(x - 5)^2 = x^2 - 10x + 25 (x + 7)^2 = x^2 + 14x + 49

Now, let's substitute these values back into the original equation:

x^2 - 10x + 25 + 2(x - 5)(x + 7) + x^2 + 14x + 49 = 0

Now, we can expand the term 2(x - 5)(x + 7):

2(x - 5)(x + 7) = 2(x^2 + 7x - 5x - 35) = 2(x^2 + 2x - 35)

Substitute it back into the equation:

x^2 - 10x + 25 + 2(x^2 + 2x - 35) + x^2 + 14x + 49 = 0

Now combine like terms:

x^2 - 10x + 25 + 2x^2 + 4x - 70 + x^2 + 14x + 49 = 0

Combine the x^2 and x terms:

4x^2 + 8x + 4 = 0

Now divide the whole equation by 4 to simplify:

x^2 + 2x + 1 = 0

Now, we have a simple quadratic equation in the form of (ax^2 + bx + c = 0). To find the solutions, we can either factor it or use the quadratic formula. In this case, let's factor it:

(x + 1)(x + 1) = 0

Now, set each factor to zero and solve for x:

1. x + 1 = 0 x = -1

Since we have a repeated root (-1 in this case), the equation (x - 5)^2 + 2(x - 5)(x + 7) + (x + 7)^2 = 0 has only one solution, which is x = -1.