高一数学在线等：设关于x的方程x^2-ax-1=0的两个实根为x1,x2,且xi<x2，函数f(x)=(2x-a)/(x^2+1)
\u5df2\u77e5\u4e8c\u6b21\u51fd\u6570f(x)=ax^2+bx+1(a,b\u2208R\uff0c\u4e14a>0),\u8bbe\u65b9\u7a0bf(x)=x\u7684\u4e24\u4e2a\u5b9e\u6839\u4e3ax1\u548cx2\u3002 \u9ad8\u4e00\u6570\u5b66\u89e3\uff1a(1)\u56e0\u4e3a\u65b9\u7a0bf\uff08x\uff09=x\u7684\u4e24\u4e2a\u5b9e\u6570\u6839\u4e3ax1\uff0cx2\uff0c
\u6240\u4ee5ax^2+\uff08b-1\uff09x+1=0\uff0c
x1+x2=(1-b)/2a,x1*x2=1/a
\u4ee4f(x)=ax^2+\uff08b-1\uff09x+1,\u5176f(x)\u7684\u5bf9\u79f0\u8f74\u4e3ax=(1-b)/2a
\u7531\u9898\u610f(x1<2<x2<4)\uff0c
f(4)=16a+4(b-1)+1>0 (i)
f(2)=4a+2(b-1)+10 (ii)(\u753b\u51fd\u6570\u56fe\uff0c\u4e00\u76ee\u4e86\u7136)
(i)+4*(ii)=-4(b-1)-3>0,\u5f97\uff0cb<1/4
(i)+2*(ii)=8a-1>0,\u5f97\uff0ca>1/8,\u53732a>1/4,\u800cbb
\u6240\u4ee5 x0+1=-b/2a+1=(2a-b)/2a>0 (a>0\u4e142a>b)
\u6240\u4ee5x0+1>0
\u6545x0\uff1e-1 \uff0c\u8bc1\u6bd5\uff01
(2)ax^2+(b-1)x+1=0
x={(1-b)\u00b1\u221a[(b-1)^2-4a]}/(2a)
\u221a[(b-1)^2-4a]\u22650,b\u22651+2\u221aa,b\u22641-2\u221aa
|x1|<2,-2<x1<2
|x2-x1|=2,x2-x1=\u00b12
\u8ba8\u8bba:
1\u3001x2-x1=2,x2>x1
x1={(1-b)-\u221a[(b-1)^2-4a]}/(2a)
x2={(1-b)+\u221a[(b-1)^2-4a]}/(2a)
x2-x1=\u221a[(b-1)^2-4a]}/a=2
-2<x1<2
-2<{(1-b)-\u221a[(b-1)^2-4a]}/(2a)<2
-2<(1-b)/(2a)-\u221a[(b-1)^2-4a]/(2a)<2
-2<(1-b)/(2a)-1<2
-1<(1-b)/(2a)<3
a>0
1+2a>b>1-6a
2\u3001x2-x1=-2,x2<x1
x1={(1-b)+\u221a[(b-1)^2-4a]}/(2a)
x2={(1-b)-\u221a[(b-1)^2-4a]}/(2a)
x2-x1=-\u221a[(b-1)^2-4a]}/a=-2,\u221a[(b-1)^2-4a]/a=2
-2<x1<2
-2<{(1-b)+\u221a[(b-1)^2-4a]}/(2a)<2
-2<(1-b)/(2a)+\u221a[(b-1)^2-4a]/(2a)<2
-2<(1-b)/(2a)+1<2
-3<(1-b)/(2a)<1
a>0
1+6a>b>1-2a
b\u7684\u53d6\u503c\u8303\u56f4\u4e3a:1\u3001x1>x2\u65f6\uff0c1+2a>b>1-6a,\u62162\u3001x1\u3008x2\u65f6\uff0c1+6a>b>1-2a
\u89e3\uff1a\u5df2\u77e5\u51fd\u6570f(x)=ax²+4x+b (a<0),
f(x)=0\u5b58\u5728\u4e24\u5b9e\u6839\u4e3ax1,x2 \u6240\u4ee54²\uff0d4ab\u22650 ==> ab\u22644
f(x)=x\u5b58\u5728\u4e24\u5b9e\u6839\u4e3a\u03b1,\u03b2. \u6240\u4ee53^2\uff0d4ab\u22650 ==> ab\u22642.25
\u82e5\u4ec5a\u4e3a\u8d1f\u6574\u6570,\u4e14f(1)=0 \u6709a+4+b=0 \u6240\u4ee5b=-4-a
\u7531\u4e8e ab=-4a-a²=4-(a+2)²\u22642.25
(a+2)²\u22651.75
\u56e0a\u4e3a\u8d1f\u6574\u6570\u6240\u4ee5 a\u2264-4 \uff08 a\u2264-4\u65f6 (a+2)²>=(-2)²=4>1.75
a=-3\u6216\uff0d1\u65f6 (a+2)^2=1 a=-2\u65f6 (a+2)²=0 )
\u53c8\u6709 x1\uff0bx2=-4/a x1\u00b7x2=b/a
==> (x1-x2)²=(x1+x2)²-4x1\u00b7x2
````````````=16/a²-4b/a=16/a²-4(-4-a)/a
````````````=16/a²+16/a+4
````````````=16[(1/a)+(1/2)]²
(x1-x2)²>=16[(1/-4)+(1/2)]²=1
(x1-x2)²<16[0+(1/2)]²=4
\u6240\u4ee5 1\u2264|x1-x2|<2.
(3)\u82e5\u03b1<1<\u03b2<2.\u8bc1\u660ex1x2<2.
f(x)-x=ax²+3x+b=a(x-\u03b1)(x-\u03b2)
\u6ce8\u610f\u5230a0 f(2)=a(2-\u03b1)(2-\u03b2)<0
\u800cf(1)=a+4+b>0 f(2)=4a+8+b<0
\u6240\u4ee5 -4-a -4/a-1>b/a>-8/a-4
\u7531-4/a-1>-8/a-4 \u5f97 -4/a<3
x1\u00b7x2=b/a<-4/a-1<3-1=2
\u89e3(1):b=2a
\u56e0\u4e3a:|\u03b1-\u03b2|=1
\u6240\u4ee5:(|\u03b1-\u03b2|)\u65b9=(a-b)\u65b9=1
\u53c8: (\u03b1+\u03b2)\u65b9 - 4\u03b1\u03b2 = (a-b)\u65b9 = 1
\u7531\u9898: \u03b1+\u03b2= b/a : \u03b1\u03b2 = -3/a \u4ee3\u5165\u4e0a\u5f0f
\u5f97\uff1aa\u65b9 \uff0b 4ab = 9
\u56e0\u4e3a\uff1aa\u65b9\uff1e0\uff1aab \uff1e0\uff0c\u4e14a;b\u4e3a\u8d1f\u6574\u6570
\u89e3\u5f97\uff1aa = -1 ; b = -2
\u89e3(2): f(x) = -x\u65b9 \uff0b 4x - 2
\u56e0\u4e3a\uff1aa + b + 4 = 0 \uff1ba\u4ec5\u4e3a\u8d1f\u6570\uff1ab\u5c31\u4e3a\u6b63\u6570
\u6240\u4ee5\uff1aa\u2264-4: b\u22650
\u53c8\u56e0\u4e3a\uff1a|x1-x2|\u65b9 \uff1d (x1 + x2)\u65b9 - 4x1x2
\u53c8\uff1ax1 + x2 = 4 / a ; x1x2 = b / a
\u6240\u4ee5\uff1a|x1-x2|\u65b9 \uff1d 4(a + 2)\u65b9 \uff0f a\u65b9
\uff1d 4\u3014(a + 2 )/a\u3015\u65b9
\uff1d 4\u30141 + 2/a \u3015\u65b9
\u53c8 :a\u2264-4 \uff1b\u6240\u4ee5\uff1a\uff081\uff0f2\uff09\u22641\uff0b 2/a\uff1c1
\u6240\u4ee5\uff1a4\u00d7\uff081\uff0f4\uff09\u22644\u30141 + 2/a \u3015\u65b9\uff1c 4
-1\u22644/a\uff1c0
\u6240\u4ee5\uff1a 1\u2264|x1-x2|<2
\u56e0\u4e3a\uff1a\u03b1\u03b2 \uff1d b/a \u4e14\uff1ax1x2 \uff1d b/a
\u53c8\uff1a\u03b1<1<\u03b2<2
\u6240\u4ee5\uff1a\u03b1\u03b2 < 2
\uff08x1+1\uff09\u00d7\uff08x2+1\uff09=x1x2+x1+x2+1=b/a+ 4/a+1< 2+ 0+1<3<7
(3)\u5f97\u8bc1
∴由一元二次方程ax2+bx+c=0(a≠0)的根的判别式Δ=b2-4ac，当Δ>0时，方程有两个不相等的实数根；两根积x1x2=c/a，x1+x2=-b/a
∴x^2-ax-1=0的a2+4>0，x1x2=-1，x1+x2= a
∴2x1x2-2-a（x1+x2）=-4-a^2
∵f(x2)-(f(x1)
=(2x2-a)/(x2^2+1)-(2x1-a)/(x1^2+1)
=[(x1-x2)(2x1x2-2-a（x1+x2）)]/[(x1^2+1)(x2^2+1)]
该式分母(x1^2+1)(x2^2+1)>0
分子中x1-x2<0，
∵x^2-ax-1=0的两个实根为x1,x2,且xi<x2
∴由一元二次方程ax2+bx+c=0(a≠0)的根的判别式Δ=b2-4ac，当Δ>0时，方程有两个不相等的实数根；两根积x1x2=c/a，x1+x2=-b/a
∴x^2-ax-1=0的a2+4>0，x1x2=-1，x1+x2= a
∴2x1x2-2-a（x1+x2）=-4-a^2=-（a2+4），-（a2+4）＜0
∴分子[(x1-x2)(2x1x2-2-a（x1+x2）>0
f(x2)-(f(x1) >0，即f(x)=(2x-a)/(x^2+1)在区间(x1,x2)上是增函数
（2）做起来好麻烦，没有时间细演算了，同（1）的道理，自己慢慢演算吧。
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