http://hk.knowledge.yahoo.com/question/question?qid=7011121800774
x - 2√(x+1) - 2 = 0
x-2 = 2√(x+1)
(x-2)^2 = [2√(x+1)]^2
x^2 - 4x + 4 = 4(x+1)
x^2 - 4x + 4 = 4x + 4
x^2 - 4x + 4 - 4x - 4 = 0
x^2 - 8x = 0
x(x-8) = 0
x = 0 或 8
驗算:
若 x = 0 ,
L.H.S =
x - 2√(x+1) - 2
= 0 - 2√(0+1) - 2
= 0 - 2√1 - 2
= 0 - 2(1) - 2
= 0 - 2 - 2
= 0 - 4
= -4
≠ R.H.S.
∴ 解 x = 0 不正確。
若 x = 8 ,
L.H.S. =
x - 2√(x+1) - 2
= 8 - 2√(8+1) - 2
= 8 - 2√9 - 2
= 8 - 2(3) - 2
= 8 - 6 - 2
= 0
= R.H.S.
∴ 解 x = 8 正確。
為什麼在解方程時 x = 0 正確,但在驗算時不正
確?