Solve: m−(m−1)2=1−(m−2)3.
Given m−(m−1)2=1−(m−2)3take LCM on both sides⇒2m−m+12=3−m+23⇒m+12=5−m3⇒3(m+1)=2(5−m) [Cross Multiplying]⇒3m+3=10−2m⇒3m+2m=10−3⇒5m=7∴m=75
Question 75
Solve the following:
m−m−12=1−m−23