Question by Huy Toàn 8A (TL)

28/07/2018 at 07:12

A = \(\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{\left(1-x\right)}{9-x^2}\)

Math 8

Mr. Bee moderators 30/07/2018 at 01:38

I think he meant to simplify the expression bro
Huy Toàn 8A (TL) selected this answer.
Tôn Thất Khắc Trịnh 28/07/2018 at 17:29

Wait what? Calculate? But you didn't even give me what x is... Well, I guess what you meant is... SOLVE

\(A=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{\left(1-x\right)}{9-x^2}=\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}+\dfrac{1-x}{x^2-9}\)

The condition for x is \(x\ne\pm3\)
\(A=\dfrac{\left(x+1\right)\left(x+3\right)}{x^2-9}+\dfrac{\left(x-1\right)\left(x-3\right)}{x^2-9}+\dfrac{1-x}{x^2-9}\)
\(\Leftrightarrow x^2+4x+3+x^2-4x+3+1-x=2x^2-x+7\)
Wait what? Are you sure you typed that question correctly?

Okaym let';s assume it's (1-x)²
Then everything will go normally, until we get:
\(\Leftrightarrow x^2+4x+3+x^2-4x+3+x^2-2x+1=3x^2-2x+7\)
Still... can't solve that...
I think you might have typed the question wrong...
Tôn Thất Khắc Trịnh 31/07/2018 at 08:27

So I've said, it's 2x²-x+7
Tôn Thất Khắc Trịnh 28/07/2018 at 17:41

How are we supposed to calculate without a given value of x?