y＝(x^2-4x+3)/(2x^2-x-1)
= (x-1)(x-3)/[(2x+1)(x-1)]
= (x-3)/(2x+1) ( x≠1)
= 1/2 - 7/[2(2x+1)]
值域 = ( -∞,-2/3) U (-2/3,+∞ )
y＝2x²-x-1分之x²-4x+3 求值域
y＝(x^2-4x+3)/(2x^2-x-1)
= (x-1)(x-3)/[(2x+1)(x-1)]
= (x-3)/(2x+1) ( x≠1)
= 1/2 - 7/[2(2x+1)]
值域 = ( -∞,-2/3) U (-2/3,+∞ )