Question

Решить уравнение:2(m+1)²+3(m-1)²-5(m+1)(m-1) = 4

Answer 1

$2(m+1)^2+3(m-1)^2-5(m-1)^2=4$

$2(m+1)^2-2(m-1)^2=4$

$(m+1)^2-(m-1)^2=2$

$(m+1+m-1)(m+1-m+1)=2$

$2m*2=2$

$m=\frac{1}{2}$

Answer 2

2(м²+1+2m)+3(m²+1-2m)-5(m²-1)=4

2m²+2+4m+3m²+3-6m-5m²+4=4

2m²+3m²-5m²+4m-6m=4-2-3-4

-2m=-5

m=5/2=2.5

Answer 3

To solve the equation, first, we need to simplify the expression:

2(m + 1)² + 3(m - 1)² - 5(m + 1)(m - 1) = 4

Let's start by expanding the squared terms:

2(m + 1)² = 2(m + 1)(m + 1) = 2(m² + 2m + 1) = 2m² + 4m + 2

3(m - 1)² = 3(m - 1)(m - 1) = 3(m² - 2m + 1) = 3m² - 6m + 3

Next, we'll expand the product term:

-5(m + 1)(m - 1) = -5(m² - 1) = -5m² + 5

Now, substitute the simplified expressions back into the original equation:

2m² + 4m + 2 + 3m² - 6m + 3 - 5m² + 5 = 4

Now, combine like terms:

(2m² + 3m² - 5m²) + (4m - 6m) + (2 + 3 + 5) = 4

-5m² - 2m + 10 = 4

To proceed, we want to get all terms on one side of the equation, so let's subtract 4 from both sides:

-5m² - 2m + 10 - 4 = 4 - 4

-5m² - 2m + 6 = 0

Now, we have a quadratic equation in the form of ax² + bx + c = 0, where a = -5, b = -2, and c = 6. To solve for m, we can use the quadratic formula:

m = (-b ± √(b² - 4ac)) / 2a

Plugging in the values:

m = (2 ± √((-2)² - 4(-5)(6))) / 2(-5) m = (2 ± √(4 + 120)) / -10 m = (2 ± √124) / -10 m = (2 ± 2√31) / -10

Now, let's find the two solutions for m:

1. m = (2 + 2√31) / -10
2. m = (2 - 2√31) / -10

Further simplification:

1. m = -(1 + √31) / 5
2. m = -(1 - √31) / 5

So, the two solutions to the equation are:

1. m ≈ -1.81
2. m ≈ 0.62