初二数学反比例函数，快帮我解解

y1=kx²,y2=m/(x-1)
y=kx²+m/(x-1)
代入（-1，3），（2，-3）解方程组
{ 3=k-m/2
{ -3=4k+m
k=0.5 , m=-5
y=0.5x²-5/(x-1)

y=0.5*2-5/(√2-1)= -5√2-4

莪觉得楼上做的是错的
因为两个正比例函数和反比例函数的K值不一样，所以就有两个K1和K2，像这样后，再向楼上的那样做，可求出K1和K2的值。就可求出Y与X之间的函数关系式啦

(1)
根据题意得:
y1=kx^2

3=k1*(-1)^2
k1=3
-3=k2*4
k2=-3/4

y2=k/(x-1)

3=k1/(-2)
k1=-6
-3=k2/(2-1)
k2=-3
即:
y1=3x^2或y1=(-3/4)x^2
y2=(-6)/(x-1)或y2=(-3)/(x-1)

y=y1+y2=3x^2+(-6)/(x-1)
或
y=y1+y2=(-3/4)x^2+(-3)/(x-1)

(2)
y=3x^2+(-6)/(x-1)
=3*2-[6/(√2-1)]
=6-[6/(√2-1)]
=6-[6(√2+1)/(√2-1)(√2+1)]
=6-(6√2+6)
=6-6√2-6
=-6√2

y=(-3/4)x^2+(-3)/(x-1)
=(-3/4)*2-[3/(√2-1)]
=(-3/2)-[3(√2+1)/(√2-1)(√2+1)]
=(-3/2)-(3√2+3)
=(-3/2)-3√2-3
=-9/2-3√2