已知(a-2)²+根号b+1/2 <等于0 求-2(3b²-2a²)+3(ab+2b²-a²)
问题描述:
已知(a-2)²+根号b+1/2 <等于0 求-2(3b²-2a²)+3(ab+2b²-a²)
答
因为(a-2)²≥0√(b+1/2)≥0所以(a-2)²+√(b+1/2)≥0要使(a-2)²+√(b+1/2) ≤0只能(a-2)²+√(b+1/2) =0得a-2=0b+1/2=0解得a=2b=-1/2-2(3b²-2a²)+3(ab+2b²-a²)=-6b²...