各位数学高手，请问这一题怎么解？
(-1/2)^-2\u8fd9\u9053\u6570\u5b66\u9898\u76ee\u600e\u4e48\u89e3\uff1f(-1/2)^-2
=1\u00f7(-1/2)²
=1\u00f7¼
=1\u00d74
=4
3\uff09\u89e3\uff1a\u56e0\u4e3a2y^2\u4e3a\u5076\u6570\uff0c\u6240\u4ee5\uff08x-y\uff09^2\u5947\u6570\uff0c\u3001
\u53c8\u56e0\uff08x-y\uff09^2\u662f\u5c0f\u4e8e27\u7684\u5b8c\u5168\u5e73\u65b9\u6570\uff0c
\u6240\u4ee5\uff08x-y\uff09^2\uff1d1\u62169\u621625
\u9010\u4e00\u9a8c\u8bc1\uff0c\uff08x-y\uff09^2\uff1d9 ,2y^2=18,y=3,x=0\u6216\uff0d6\uff1by=\uff0d3,x=0\u62166\uff0c;
\uff08x-y\uff09^2\uff1d25\uff0c2y^2\uff1d2.y=1\u65f6x=\uff0d6\u62164;y=\uff0d1\u65f6x=\uff0d4\u62166.
\u6240\u4ee5x=\uff0d6;\uff0d4;0;4;6.
(\u5171\u6709\u516b\u7ec4\u7ed3\u679c\uff1ax=4\uff0cy=1\uff1bx=\uff0d6\uff0cy=1\uff1bx=6\uff0cy=\uff0d1\uff1bx=\uff0d4\uff0cy=\uff0d1\uff1bx=0\uff0cy=3\uff1bx=0\uff0cy=\uff0d3\uff1bx=6\uff0cy=0\uff1bx=\uff0d6\uff0cy=\uff0d3)
f(x)在x=3处的左极限为9a-3b+3，右极限为6-a+b
由已知：4=4a-2b+3
9a-2b+3=6-a+b
解之得a=-3/2，b=-7/2
因为连续，所以
f(2)=lim....=4(求极限用洛必达法则)=a(4)-b(2)+3
f(3)=a(9)-b(3)+3=2(3)-a+b
可省{10a-4b=3 8a-4b=2}
a=0.5
b=0.5
希望对你有帮助.
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