已知(a-2)²+根号b+1/2 <等于0 求-2(3b²-2a²)+3(ab+2b²-a²)
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已知(a-2)²+根号b+1/2 <等于0 求-2(3b²-2a²)+3(ab+2b²-a²)
已知(a-2)²+根号b+1/2 <等于0 求-2(3b²-2a²)+3(ab+2b²-a²)
已知(a-2)²+根号b+1/2 <等于0 求-2(3b²-2a²)+3(ab+2b²-a²)
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=0
b+1/2=0
解得a=2
b=-1/2
-2(3b²-2a²)+3(ab+2b²-a²)
=-6b²+4a²+3ab+6b²-3a²
=a²+3ab
=2²+3×2×(-1/2)
=4-3
=1
答案:1
(a-2)²+√(b+1/2 )<=0 => 小於0不可能 =>(a-2)²+√(b+1/2 )=0 => a=2 ;b= -1/2
-2(3b²-2a²)+3(ab+2b²-a²)=-6b²+4a²+3ab+6b²-3a²=a² +3ab =4+3*2*(-1/2)=1
(a-2)²+根号b+1/2 <等于0 得a=2 b=1/2 代入-2(3b²-2a²)+3(ab+2b²-a²)=1
由已知: (a-2)²+√(b+1/2)<=0, 该式<0不可能!只能=0,从而 解出:a=2、b=-1/2;因此: -2(3b²-2a²)+3(ab+2b²-a²) =-6b²+4a²+3ab+6b²-3a² =a²+3ab =4-3 =1