Consider the equation x^2-x-2/x+1=x-2
When x = 0,1,2,...,L.H.S.=R.H.S.(Check it yourself.)
Can we say that it is an identity?Explain
Open-ended Question
2009-06-30 6:06 am
回答 (3)
2009-06-30 7:33 am
Your equation should be (x^2 - x - 2) / (x + 1) = x - 2.
It is not an identity since the L.H.S. is not in the simplest form.
L.H.S. = (x^2 - x - 2) / (x + 1) = [(x - 2)(x + 1)] / (x + 1) = x - 2
Examples for identity are :
a^2 - b^2 = (a + b) (a - b)
(a + b)^2 = a^2 + 2ab + b^2
It is not an identity since the L.H.S. is not in the simplest form.
L.H.S. = (x^2 - x - 2) / (x + 1) = [(x - 2)(x + 1)] / (x + 1) = x - 2
Examples for identity are :
a^2 - b^2 = (a + b) (a - b)
(a + b)^2 = a^2 + 2ab + b^2
2009-06-30 6:26 am
when x=-1 then x+1=0 so it isn't an identity
but in R－{1} it is an identity.
but in R－{1} it is an identity.
2009-06-30 6:22 am
代入0已經不對了
收錄日期: 2021-04-13 16:42:02
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