f(x) = 1/ (2x − 5)^2(3x − 1)based at b = 0. I tried partial fractions and I think I'm doing it wrong. Thanks in advance.
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Originally Posted by khuezy f(x) = 1/ (2x − 5)^2(3x − 1)based at b = 0. I tried partial fractions and I think I'm doing it wrong. [snip] Then please show your working.
1 = A(2x-5)(3x-1) + B(3x-1) + C(2x-5)(2x-5) x=5/2 1= B(3(5/2)-1) B=2/13???? That doesn't seem right.
Originally Posted by khuezy 1 = A(2x-5)(3x-1) + B(3x-1) + C(2x-5)(2x-5) x=5/2 1= B(3(5/2)-1) B=2/13???? That doesn't seem right. Sorry but I cannot follow this working. Please show all the working, set out clearly. If you expect to get a clear reply to your problem, then surely it's reasonable that you clearly show all your working when requested.
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