设fx=ax²+(b-1)x-a-ab,方程fx=0的两根是-2,0.
\u82e5\u65b9\u7a0bax²=b\ufe59ab>0\ufe5ax²=b/a
\u56e0\u4e3a\u4e24\u4e2a\u6839\u5206\u522b\u662fm+1\u548c2m-4
\u6240\u4ee5
m+1+2m-4=0
3m=3
m=1
\u6839\u4e3a2\u6216-2
\u6240\u4ee5
b/a=2²=4
ax^2+by^2=ab \u4e24\u8fb9\u540c\u9664\u4ee5 ab \u5f97 x^2/b+y^2/a=1 \u3002
\uff081\uff09\u5982\u679c a>0\uff0cb>0 \uff0c\u8868\u793a\u692d\u5706\u6216\u5706\uff1b
\uff082\uff09\u5982\u679c a0 \u6216 a>0\uff0cb<0 \uff0c\u8868\u793a\u53cc\u66f2\u7ebf \uff1b
\uff083\uff09\u5982\u679c a<0\uff0cb<0 \uff0c\u4e0d\u8868\u793a\u4efb\u4f55\u66f2\u7ebf\u3002
ax+by+c=0 \u8868\u793a\u76f4\u7ebf \u3002
所以-（b-1)/a=-2+0
-a-ab=(-2）*0=0
解得：a=-1,b=-1
所以ab=1
2，f(x)=-x^2-2x,g(x)=(-x^2-2x)/(x^2+2)-2,(x∈[2,4])
=[-3（x^2+2)-2(x-1)]/(x^2+2)
=-3-2(x-1)/(x^2+2)
=-3-2(x-1)/[(x-1)^2+2(x-1)+3]
=-2/[(x-1)+3/(x-1)+2]
令r(x)=(x-1)+3/(x-1)+2,x∈[2,4]
所以r(x)>=2√[(x-1)*3/(x-1)] +2=2+2√3，当x=1+√3时等号成立，所以r(x)在x∈[2,1+√3],上单调减，在x∈(1+√3,4]上单调增，而r(2)=7,r(4)=6，所以r(x)max=7,r(x)min=2+2√3
g(x)min=-2/(2+2√3）=（1-√3）/2,
g(x）max=-2/7
解,1，因为f(x）=ax^2+(b-1)x-a-ab=0的两根是-2,0,所以a≠0
所以-（b-1)/a=-2+0
-a-ab=(-2）*0=0
解得：a=-1,b=-1
所以ab=1
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