y=(x^2-1)/(x^2-5x+4)

求值域
求过程

先求定义域：x^2-5x+4<>0,(x-1)(x-4)<>0,x<>1且x<>4
然后求值域
y=(x^2-1)/(x^2-5x+4)=(x+1)(x-1)/(x-1)(x-4)=(x+1)/(x-4)=1+5/(x-4)
由于x<>1，所以5/(x-4)<>5/(1-4)=-5/3,5/(x-4)<>0
y=1+5/(x-4),y<>-2/3,y<>1

2. x^4+y^4+(x+y)^4 =(x^2+y^2)^2-2x^2y^2+(x+y)^4 =[(x+y)^2-2xy]^2-2x^2y^2+(x+y)^4 =[(x+y)^2]^2-4xy(x+y)^2+4x^2y^2-2x^2y^2+(x+y)^4 =2(x+y)^4-4xy(x+y)^2+2x^2y^2 =2[(x+y)^4-2xy(x+y)^2+(xy)^2] =2[(x+y)^2-xy]^2 =2(x^2+xy+y^2)^2.

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