若/2a+1/+4a2-4ab+b2=0求a2{a-b}-b2{b-a]的值

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若/2a+1/+4a2-4ab+b2=0求a2{a-b}-b2{b-a]的值

若/2a+1/+4a2-4ab+b2=0求a2{a-b}-b2{b-a]的值
若/2a+1/+4a2-4ab+b2=0求a2{a-b}-b2{b-a]的值

若/2a+1/+4a2-4ab+b2=0求a2{a-b}-b2{b-a]的值
/2a+1/+4a^2-4ab+b^2=0
/2a+1/+(2a-b)^2=0
/2a+1/=0,(2a-b)^2=0
a=-1/2,b=2a
b=2*(-1/2)=-1
a^2(a-b)-b^2(b-a)
=a^2(a-b)+b^2(a-b)
=(a^2+b^2)(a-b)
=[(-1/2)^2+(-1)^2][-1/2-(-1)]
=(1/4+1)(-1/2+1)
=5/4*1/2
=5/8

/2a+1/+4a2-4ab+b2=0
|2a+1|+(2a-b)²=0
∴2a+1=0
2a-b=0
∴a=-1/2
b=-1
∴a2{a-b}-b2{b-a]
=a²(a-b)+b²(a-b)
=(a-b)(a²+b²)
=(-1/2+1)(1/4+1)
=5/8