F.3 Mathematics (3 questions)

A. (x^2 + y^2)^1/2 = x+y
B. (x^1/2 + y^1/2)^2 = x+y
C. (x^2 * y^2)^1/2 = xy
D. (2x^2)^3 = 2x^6
E. (2x^3)^2 = 4x^5
For (A),
L.H.S. = (x^2 + y^2)^1/2
= √(x^2 + y^2)
= √[(x + y)^2 - 2xy]
≠ R.H.S.
∴(A) is wrong.
For (B),
L.H.S. = (x^1/2 + y^1/2)^2
= (√x + √y)^2
= x + y + 2√(xy)
≠ R.H.S.
∴(B) is wrong.
For (C),
L.H.S. = (x^2 * y^2)^1/2 = xy
= √(x^2 * y^2)
= √x^2 * √y^2
= xy
= R.H.S.
∴(C) is correct.
For (D),
L.H.S. = (2x^2)^3
= 2^3 * (x^2)^3
= 8x^6
≠ R.H.S.
∴(D) is wrong.
For (E),
L.H.S. = (2x^3)^2
= 2^2 * (x^3)^2
= 4x^6
≠ R.H.S.
∴(E) is wrong.
9.Simplify (a^1/2 + b^1/2)(a^1/2 - b^1/2)
(a^1/2 + b^1/2)(a^1/2 - b^1/2)
=(√a + √b)(√a - √b)
=√a^2 - √b^2
=a - b
13. If x^4 = 16a^4 * b^12 ,find the value of x.
x^4 = (2a)^4 * (b^3)^4
x = 4√[(2a)^4 * (b^3)^4]
= 2ab^3

2009-09-24 23:23:23 補充：
For (A),
L.H.S. = (x^2 + y^2)^1/2
= √(x^2 + y^2)
= √[(x + y)^2 - 2xy]
≠ R.H.S.
∴(A) is wrong.

∵(x + y)^2 = x^2 + 2xy + y^2
∴x^2 + y^2 = (x+y)^2 - 2xy

如果你都係唔明，可以掉數計...
(x+y)^2 - 2xy
=x^2 + 2xy + y^2 - 2xy
=x^2 + y^2

2009-09-24 23:29:51 補充：
Q13... 佢show唔倒個+/- o既符號出o黎... (±)
即係例如...
設 x^2 = 9
唔係 x = 3 咁簡單...
而 3^2 = 9係o岩.... 但係(-3)^2都係 = 9...
所以，x^2 = 9, x = +/- √9 = +/- 3
所以:
13. If x^4 = 16a^4 * b^12 ,find the value of x.
x^4 = (2a)^4 * (b^3)^4
x = +/- (^)4√[(2a)^4 * (b^3)^4]
= +/- 2ab^3

8.Which of the following is correct?

C. (x^2 * y^2)^1/2 = xy

9.Simplify (a^1/2 + b^1/2)(a^1/2 - b^1/2)
= a-b 因為 (a+b)(a-b)=a^2 - b^2

13. If x^4 = 16a^4 * b^12 ,find the value of x.
x^4 = (2a)^4 * (b^3)^4
x=2ab^3 因為兩邊都除四次.......