Question

A^2-b^2=6(a-2)^2-(b-2)^2=18найти (a+b)

Answer 1

{а² - b² = 6

{(a-2)² - (b-2)² = 18

найти (a+b)

Решение

Применим формулу разности квадратов: а² - b²=(а - b)(а + b)

{(а - b)(a + b) = 6

{(a-2-(b-2))·(a-2+ (b-2)) = 18

{(а - b)(a + b) = 6

{(a-b))·(a+b-4)) = 18

{a - b = 6 /(a+b)

{6/(a+b) · (a+b-4) = 18

6/(a+b) · (a+b-4) = 18

6/(a+b) = 18/(a+b-4)

6 · (a+b-4) = 18 ·(a+b)

(a+b-4) = 3(a+b)

(a+b) - 4 = 3(a+b)

(a+b)-3(a+b) = 4

-2(a+b) = 4

(a+b) = 4 : (-2)

a+b = - 2

Answer 2

Let's solve the given equations step by step to find the value of (a + b).

Equation 1: A^2 - b^2 = 6(a - 2)^2 - (b - 2)^2 Equation 2: 6(a - 2)^2 - (b - 2)^2 = 18

We can simplify Equation 1 by expanding the squares:

(A + b)(A - b) = 6(a^2 - 4a + 4) - (b^2 - 4b + 4) A^2 - b^2 = 6a^2 - 24a + 24 - b^2 + 4b - 4 0 = 6a^2 - 24a + 4b - 20

Now we substitute Equation 2 into the simplified Equation 1:

6(a - 2)^2 - (b - 2)^2 = 18 6(a^2 - 4a + 4) - (b^2 - 4b + 4) = 18 6a^2 - 24a + 24 - b^2 + 4b - 4 = 18 6a^2 - 24a - b^2 + 4b + 6 = 0

Comparing the coefficients of the simplified Equation 1 and Equation 2, we have:

6a^2 - 24a + 4b - 20 = 6a^2 - 24a - b^2 + 4b + 6

Simplifying further, we get:

5b - 26 = 0 b = 26/5

Substituting the value of b back into the equation, we can solve for a:

6a^2 - 24a + 4(26/5) - 20 = 0 30a^2 - 120a + 104 - 100 = 0 30a^2 - 120a + 4 = 0 15a^2 - 60a + 2 = 0

Using the quadratic formula, we can find the solutions for a:

a = (-(-60) ± sqrt((-60)^2 - 4 * 15 * 2))/(2 * 15) a = (60 ± sqrt(3600 - 120))/(30) a = (60 ± sqrt(3480))/(30) a = (60 ± sqrt(3480))/(30) a = (60 ± 2*sqrt(870))/(30) a = 2(30 ± sqrt(870))/(30) a = 2(1 ± sqrt(870)/15)

Now we can substitute the values of a and b into (a + b):

(a + b) = [2(1 + sqrt(870)/15) + 26/5] or [2(1 - sqrt(870)/15) + 26/5]

Calculating both possibilities, we get:

(a + b) = 2 + 2sqrt(870)/15 + 26/5 or 2 - 2sqrt(870)/15 + 26/5

Hence, the value of (a + b) is either 2 + 2sqrt(870)/15 + 26/5 or 2 - 2sqrt(870)/15 + 26/5, depending on the values of a and b that satisfy the equations.