Question

|a+1|+(b-3)^2=0,求代数式[2a+b)^2-(2a+b)(2a-b)-6b]÷(-2b)

Answer 1

|a+1|+(b-3)^2=0

|a+1|=0且(b-3)^2=0

解得a=-1,b=3

代入[2a+b)^2-(2a+b)(2a-b)-6b]÷(-2b)=（1-1*-5-6*3）=-12

已知x^2-6x+8=0即（x-2）（x-4）=0解得x=2或x=4

将x=2代入[2x(x-3)^2]^3÷(x-3)^2=[2*2（-1）^2]^3/(2-3)^2=64

将x=4代入[2x(x-3)^2]^3÷(x-3)^2=[2*4（1）^2]^3/(4-3)^2=512