6.\(m-\frac{m-1}2=1-\frac{m-2}3\)

6.\(m-\frac{m-1}2=1-\frac{m-2}3\) Given that,\(m-\frac{m-1}2=1-\frac{m-2}3\)
LCM of denominators 3 and 4=6

So, \(m×6-\frac{m-1}2×6=1×6-\frac{m-2}3×6 \quad \)[Multiplying both the sides by 6]

\(\Rightarrow 6m – (m – 1) × 3 = 6 – (m – 2) × 2\)

\(\Rightarrow 6m – 3m + 3 = 6 – 2m + 4 \quad \)[Solving the brackets]

\(\Rightarrow 3m + 3 = 10 – 2m \)

\(\Rightarrow 3m + 2m = 10 – 3 \)[Transposing 2m to LHS and 3 to RHS]

\(\Rightarrow 5m = 7\)

\(\Rightarrow m = \frac { 7 }{ 5 } \)[Transposing 5 to RHS]

Hence \(m = \frac { 7 }{ 5 } \) is the required solution.