有關 **form 2** 數學的問題 ~~~識先好來!!!!亂答即告~~~ @_@

已知x=1/(n+1)和y=n/(n-1)。

(a) (i) 試以 x 表示 n 。

x = 1/(n+1)
x(n+1) = 1
xn + x = 1
xn = 1-x
n = (1-x)/x

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(ii) 試以 y 表示 n 。

y = n/(n-1)
y(n-1) = n
ny - y = n
ny - n = y
n(y-1) = y
n = y/(y-1)

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(b) (i) 試以 x 表示 y 。

從 (a)(i) 及 (ii)

n = (1-x)/x
n = y/(y-1)

所以
(1-x)/x = y/(y-1)
(y-1)(1-x) = xy
y - xy - 1 + x = xy
y - 2xy - 1 + x = 0
y(1-2x) = 1-x
y = (1-x)/(1-2x)

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(ii) 試以 y 表示 x 。

從 (a)(i) 及 (ii)

n = (1-x)/x
n = y/(y-1)

所以
(1-x)/x = y/(y-1)
(y-1)(1-x) = xy
y - xy - 1 + x = xy
y - 2xy - 1 + x = 0
x(1-2y) = 1-y
x = (1-y)/(1-2y)

(a) (i) x=1/(n+1)
1/x=n+1
(1/x)-1=n
n=(1/x)-1 OR (1-x)/x
(ii) y=n/(n-1)
y(n-1)=n
yn-y=n
y=yn-n
y=n(y-1)
n=y/(y-1)

(b) (i) because n=(1-x)/x and n=y/(y-1)
therefore (1-x)/x=y/(y-1)
(1-x)(y-1)=xy
y-1-xy+x=xy
x-1=xy+xy-y
x-1=y(2x-1)
y=(x-1)/(2x-1)
(ii) y=(x-1)/(2x-1)
y(2x-1)=x-1
2xy-y=x-1
x-2xy=1-y
x(1-2y)=1-y
x=(1-y)/(1-2y)