Question

若|2a-b|+(3b-2)平方=0,求1/2(a-b)+1/4(a+b)+(a+b)/3-(a+b)/6的值

Answer 1

因为：|2a-b|+(3b-2)=0 所以得：2a-b=0；3b-2=0 解得：a=1/3,b=2/3；代入下式 1/2(a-b)+1/4(a+b)+(a+b)/3-(a+b)/6 =a/2-b/2+a/4+b/4+a/3+b/3-a/6-b/6 =a(1/2+1/4+1/3-1/6)+b(-1/2+1/4+1/3-1/6) =ax11/12-b/12 =(1/3)x(11/12)-(2/3)x(1/12) =1/4