Expand
(i) ` (2a-5b-7c)^2`
(ii) ` ( -3a+4b-5c)^2`
(iii) `((1)/(2)a-(1)/(4)b+2)^2`

Answer

Let's expand the given expressions step by step.

### (i) Expand \( (2a - 5b - 7c)^2 \)

1. **Identify the terms**: Here, \( a = 2a \), \( b = -5b \), and \( c = -7c \).
2. **Apply the identity**: Use the identity \( (A + B + C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA \).
3. **Calculate each term**:
- \( A^2 = (2a)^2 = 4a^2 \)
- \( B^2 = (-5b)^2 = 25b^2 \)
- \( C^2 = (-7c)^2 = 49c^2 \)
- \( 2AB = 2(2a)(-5b) = -20ab \)
- \( 2BC = 2(-5b)(-7c) = 70bc \)
- \( 2CA = 2(-7c)(2a) = -28ac \)
4. **Combine the results**:
\[ (2a - 5b - 7c)^2 = 4a^2 + 25b^2 + 49c^2 - 20ab + 70bc - 28ac \]

### (ii) Expand \( (-3a + 4b - 5c)^2 \)

1. **Identify the terms**: Here, \( A = -3a \), \( B = 4b \), and \( C = -5c \).
2. **Apply the identity**: Use the same identity as above.
3. **Calculate each term**:
- \( A^2 = (-3a)^2 = 9a^2 \)
- \( B^2 = (4b)^2 = 16b^2 \)
- \( C^2 = (-5c)^2 = 25c^2 \)
- \( 2AB = 2(-3a)(4b) = -24ab \)
- \( 2BC = 2(4b)(-5c) = -40bc \)
- \( 2CA = 2(-5c)(-3a) = 30ac \)
4. **Combine the results**:
\[ (-3a + 4b - 5c)^2 = 9a^2 + 16b^2 + 25c^2 - 24ab - 40bc + 30ac \]

### (iii) Expand \( \left( \frac{1}{2}a - \frac{1}{4}b + 2 \right)^2 \)

1. **Identify the terms**: Here, \( A = \frac{1}{2}a \), \( B = -\frac{1}{4}b \), and \( C = 2 \).
2. **Apply the identity**: Again, use the same identity.
3. **Calculate each term**:
- \( A^2 = \left(\frac{1}{2}a\right)^2 = \frac{1}{4}a^2 \)
- \( B^2 = \left(-\frac{1}{4}b\right)^2 = \frac{1}{16}b^2 \)
- \( C^2 = (2)^2 = 4 \)
- \( 2AB = 2\left(\frac{1}{2}a\right)\left(-\frac{1}{4}b\right) = -\frac{1}{4}ab \)
- \( 2BC = 2\left(-\frac{1}{4}b\right)(2) = -\frac{1}{4}b \)
- \( 2CA = 2(2)\left(\frac{1}{2}a\right) = 2a \)
4. **Combine the results**:
\[ \left( \frac{1}{2}a - \frac{1}{4}b + 2 \right)^2 = \frac{1}{4}a^2 + \frac{1}{16}b^2 + 4 - \frac{1}{4}ab - \frac{1}{2}b + 2a \]