Question

2x/x-4+1/x-2+2/x^2-6x+8=0

Answer 1

By properly writing the question,

2x/(x-4) + 1/(x-2) + [2/(x²-6x+8)] = 0

Let, A = 2/(x²-6x+8).

By factorising the denominator of 'A',
we get,
x²-6x+8 = (x-2)(x-4)

So the equation becomes,
2x/(x-4) + 1/(x-2) + [2/(x-2)(x-4)] = 0

by taking 1/(x-2)(x-4) as a common term, we get,

1/(x-2)(x-4) [2x(x-2) + (x-4) + 2] = 0

multiplying (x-2)(x-4) on both sides, we get,

[2x(x-2) + (x-4) + 2] = 0

on simplifying,

2x² - 4x + x - 4 + 2 = 0

2x² - 3x - 2 = 0

on factorising,
(2x - 4) (2x + 1) = 0

so,
_____________
| x = 2 and -1/2 |
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Yes. Girls exist who are good in Mathematics.

- WonderGirl