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secondary 2 | Maths
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Date Posted: 4 years ago
Views: 341
(a)
note:
1. x² + x = x(x+1)
2. x² - 1 = (x-1)(x+1)
so
(x+3)/(x² + x) + (x-3)/(x² - 1)
= (x+3)(x-1)/[x(x+1)(x-1)]
+ (x-3)(x)/[x(x+1)(x-1)]
= (x² + 2x -3)/[x(x+1)(x-1)]
+ (x² - 3x)/[x(x+1)(x-1)]
= (2x² - x -3)/[x(x+1)(x-1)]
= (2x - 3)(x + 1)/[x(x+1)(x-1)]
= (2x - 3)/[x(x-1)]
(b)
x/3 = √[(y+2)/(y-2)]
square both sides,
x²/9 = (y+2)/(y-2)
x²(y-2) = 9(y+2)
x²y - 2x² = 9y + 18
x²y - 9y = 2x² + 18
y(x² - 9) = 2x² + 18
y = (2x² + 18)/(x² - 9)
note:
1. x² + x = x(x+1)
2. x² - 1 = (x-1)(x+1)
so
(x+3)/(x² + x) + (x-3)/(x² - 1)
= (x+3)(x-1)/[x(x+1)(x-1)]
+ (x-3)(x)/[x(x+1)(x-1)]
= (x² + 2x -3)/[x(x+1)(x-1)]
+ (x² - 3x)/[x(x+1)(x-1)]
= (2x² - x -3)/[x(x+1)(x-1)]
= (2x - 3)(x + 1)/[x(x+1)(x-1)]
= (2x - 3)/[x(x-1)]
(b)
x/3 = √[(y+2)/(y-2)]
square both sides,
x²/9 = (y+2)/(y-2)
x²(y-2) = 9(y+2)
x²y - 2x² = 9y + 18
x²y - 9y = 2x² + 18
y(x² - 9) = 2x² + 18
y = (2x² + 18)/(x² - 9)
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Date Posted:
4 years ago
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