ratio

a ) Let a / b = c / d = k.

So a = bk, c = dk

L.H.S. = a / ( a + b )

= bk / ( bk + b )

= k / ( k + 1 )

R.H.S. = c / ( c + d )

= dk / ( dk + d )

= k / ( k + 1 )

= L.H.S.

b ) ( 2x - 5 ) / ( x^2-2x+5 ) = ( 3x + 2 ) / ( x^2-3x-2 )

( 2x - 5 ) / ( 2x - 5 + x^2 - 2x + 5 ) = ( 3x + 2 ) / ( 3x + 2 - x^2 - 3x - 2 )

( 2x - 5 ) / x^2 = ( 3x + 2 ) / x^2

x^2 ( 2x - 5 - 3x - 2 ) = 0

x^2 ( x + 7 ) = 0

So x = 0 or -7




2007-07-11 20:55:49 補充：
For part b, let a = 2x - 5, b = x^2 - 2x + 5, c = 3x + 2 and d = x^2 - 3x - 2 and so by using a, x can be solved. Also ,sorry for a minor tying mistake, the yellow part should be ( 2x - 5 ) / ( 2x - 5 + x^2 - 2x + 5 ) = ( 3x + 2 ) / ( 3x + 2 + x^2 - 3x - 2 ).

a)
a / b = c / d
b / a = d / c
b / a + 1 = d / c + 1
(a+b) / a = (c+d) / c
a / (a+b) = c / (c+d), as long as the denominator not equal to 0.

b)
Let a = 2x - 5, b = x² - a, c = 3x + 2, d = x² - c.
2x + 53x + 2
-------------- = -----------------
x² - 2x + 5x² - 3x - 2
ac
------- = ---------
x² - ax² - c

Case (i): x² ≠ 0 ==> x ≠ 0
(This is to be considered because if x = 0 the equality below fails)
a / x² = c / x² (by a)
2x - 5 = 3x + 2
x = -7

Case (ii): x = 0
Put x = 0 into original equation, we have LHS = 0 = RHS. Hence x = 0 is an answer.

Hence, in conclusion, x = 0 or -7.