高中数学竞赛题
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1)已知x>0,y>0，且x+9y+xy=16，则y<>-1;x=(16-9y)/(1+y)=-9+25/(1+y);
xy=-9y+25y/(1+y)=-9y+25-25/(1+y)=-9(y+1)+34-25/(1+y)<=34-2*3*5=4;
则xy取最大值4时，9(y+1)=25/(y+1),y+1=3*5,y=14,x=4/14=2/7;x/2+3y=1/7+42;
2)
已知关于x的一元二次不等式ax^2-4x+b>0的解集为{x丨x≠2/a}，且a>b，则（a+b）^2/(a+b)的最小值为？
ax^2-4x+b>0,a<>0,否则不等式的解集为x<b/4和已知矛盾;
所以a(x-2/a)^2>4/a-b;
如果a<0,则不等式的解集为-√(4/a^2-b/a)+2/a<x<√(4/a^2-b/a)+2/a和已知矛盾;
所以a>0,则x>√(4/a^2-b/a)+2/a或x<-√(4/a^2-b/a)+2/a;
所以4/a^2-b/a=0,b=4/a;
则（a+b）^2/(a+b)=a+b=a+4/a>=2*2=4;
3)
已知函数f(x)=(1/2)^x的反函数为g(x)=-lnx/ln2;
g(a)+g(b)=-ln(ab)/ln2=3,ab=1/8,b=1/(8a);
1/a+1/(2b)=1/a+4a>=2*2=4
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绛旓細1)宸茬煡x>0,y>0锛屼笖x+9y+xy=16锛屽垯y<>-1;x=(16-9y)/(1+y)=-9+25/(1+y);xy=-9y+25y/(1+y)=-9y+25-25/(1+y)=-9(y+1)+34-25/(1+y)<=34-2*3*5=4;鍒檟y鍙栨渶澶у4鏃讹紝9(y+1)=25/(y+1),y+1=3*5,y=14,x=4/14=2/7;x/2+3y=1/7+42;2)宸茬煡鍏充簬...
