Question

-6(2+m)(2-m)+(5-m)^2-m(7m-1)=0

Answer 1

$- 6(4 - {m}^{2} ) +2 5 - 10m + {m}^{2} - 7 {m}^{2} + m = 0 \\ - 24 + 6 {m}^{2} + 25 - 9m - 6 {m}^{2} = 0 \\ 9m = 1 \\ m = \frac{1}{9}$

Answer 2

Let's solve the given equation step-by-step:

1. Expand the terms:

-6(2+m)(2-m) + (5-m)^2 - m(7m-1) = 0

2. Expand each part:

-6(4 - m^2) + (5 - m)^2 - (7m^2 - m) = 0

3. Distribute the -6:

-24 + 6m^2 + (5 - m)^2 - (7m^2 - m) = 0

4. Expand the squared term:

-24 + 6m^2 + (5 - m)(5 - m) - (7m^2 - m) = 0

5. Expand the remaining squared term and simplify:

-24 + 6m^2 + (25 - 10m + m^2) - (7m^2 - m) = 0

6. Combine like terms:

-24 + 6m^2 + 25 - 10m + m^2 - 7m^2 + m = 0

7. Simplify further:

6m^2 - 7m^2 + m^2 - 10m + m - 24 + 25 = 0

8. Combine the m^2 terms and the m terms:

0 = 0

The equation simplifies to 0 = 0, which is an identity. This means that the given equation is true for all real values of 'm'. In other words, the equation is satisfied by any value of 'm', and there is no specific solution for 'm' in this case.