设函数y1(x),y2(x),y(3x)都是线性方程y-搜答案-神马作文网

A(x1,1/2+log2(x1/(1-x1))B(x2,1/2+log2(x2/(1-x2))OM=1/2*(OA+OB)=1/2*(x1+x2,1+log2(x1x2/(1-x1-x2+x1x2)

（1）根据题意,可这样设：y1=k1xy2=k2/x(k1,k2为常数）（2）y=y1+y2=k1x+k2/x将x=1,y=-3；x=2,y=0代入y,得k1+k2=-32k1+k2/2=0联立解得k

第一问,将原式化简为1-（1/2^(x-1/2)+1）然后x1+x2=1,所以y1+y2=2-(（1/2^(x1-1/2)+1）+（1/2^(x2-1/2)+1）)=2-(（1/2^(x1-1/2)+

y1>y2-x+3>3x-4-x-3x>-4-3-4x>-7x所以当xy2

（1)当a=0,解得a-2综上,a

1)由题意有a^3x+1=a^（-2）x化简有(a^（-2）-a^3）x=1∵a＞0且a≠1,∴a^（-2）-a^3≠0∴当x=1/(a^（-2）-a^3)时,y1=y22)由题意,a^3x+1＞a^

y1与x二次方成反比例y1=a/x^2y2与x+2成正比例y2=b(x+2)y=y1-y2=a/x^2-b(x+2)根据题意得：9=a-b(1+2)5=a-b(-1+2)解之得：a=3,b=-2y与x

由题意得：当x=a时y1=5,y2=25y1+y2=x^2+16x+135+25=a^2+16a+13a^2+16a-17=0（a+17）（a-1）=0a>0所以a=-17（舍去）,a=1x=a=1时

S(n)=(n-1)/2+log(1/n)-log(n-1)/n)+log(2/n)-log(n-2)/n)+log(3/n)-log(n-3)/n)+.log(n-1/n)-log(1)/n).显然

y1>y2-x+3>3x-44x

由y1-y2=x^2-2x-3=(x-3)(x+1),得1)当x>3orxy22)当-1

xy975765506：-x+3=-3x+123x-x=12-32x=9x=9/2∴y=-9/2+3=-3/2∴交点坐标是(9/2,-3/2)y1>y2时,-x+3>-3x+123x-x>12-32x

解由当x=2时,y1+y2=-1得2k1+2k2=-1即k1+k2=-1/2.①当x=3时,y1-y2=12得3k1-3k2=12即k1-k2=4.②由（1）与（2）联立解得k1=7/4,k2=-9/

先看看行不x=-100:.1:100;y1=f(x);y2=g(x);plot(y1,y2);

已知y1是关于x的正比例函数,y2是关于x的反比例函数,并且当自变量x取1,y1=y2；当自变量x取2时,y1-y2=9.求y1和y2的关系式.设y1=ax,a>0；y2=b/x,b>0当自变量x取1

y1=-x+3,y2=3x-5因为y1>y2所以-x+3>3x-54x

当y1＞y2时，2x-6＞-5x+1，即7x＞7，解得：x＞1，因此当x＞1时，y1＞y2．

y=(1-C1-C2)y1(x)+C1y2(x)+C2y3(x)即y=y1(x)+C1*[y2(x)-y1(x)]+C2*[y3(x)-y1(x)]而y1(x),y2(x),y3(x)都是线性方程y'

若y1=y2那么a^(2x-7)=a^(4x-1)∴2x-7=4x-1解得:x=-3∴x=-3时,y1=y2若y1>y2那么a^(2x-7)>a^(4x-1)当a>1时,y=a^x为增函数∴2x-7>

y1=y22x+1=5x+16-3x=15x=-5y1>y22x+1>5x+16-3x>15x>-5y1