Question

已知丨a-2丨+（b+1）²+丨c丨=0，求a²-2ab-5a²+12ac+3ab-c²-8ac-2a平方？？ ？？？

Answer 1

丨a-2丨+（b+1）²+丨c丨=0，
丨a-2丨>=0,（b+1）²>=0,丨c丨>=0，
三个非负数之和等于0，则三个数全为0
a-2=b+1=c=0
a=2,b=-1,c=0
a²-2ab-5a²+12ac+3ab-c²-8ac-2a²
=-6a²+ab+4ac
=a(-6a+b+4c)
=2×(-12-1)
=-26

Answer 2

丨a-2丨+（b+1）²+丨c丨=0，
a-2=b+1=c=0
a=2,b=-1,c=0
a²-2ab-5a²+12ac+3ab-c²-8ac-2a²
=a(-6a+b+4c)
=2×(-12-1)
=-26