Question

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

Answer 1

解：由y=(x²-2x+1)/(x+1)，可知
x²-2x+1=y(x+1)
即x²-(2+y)x+1-y=0
易知，只有存在x时，才有y，即上述方程必有实根
∴△=[-(2+y)]²-4·1·(1-y)=y²+4y+4-4+4y=y(y+8)≥0
∴y≤-8或y≥0
∴原函数的值域为{yly≤-8或y≥0}

Answer 2

y=（x^2-2x+1)/（x^2-1)÷(x^2-x)/(x+1)-1/x+1
=(x-1)^2/(x-1)(x+1)÷x(x-1)/(x+1)-1/(x+1)
=(x-1)/(x+1)*(x+1)/x(x-1)-1/(x+1)
=1/x-1/x+1
=1
所以,在等号右边代数式有意义的条件下，不论x为何值，y的值都不变,始终是1