M = \(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\)
= \(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\)
= \(\dfrac{\left(x+1\right)^2+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{10}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{5}{\left(x-1\right)\left(x+1\right)}\)
b,
Để M = 5
<=> \(\dfrac{5}{\left(x-1\right)\left(x+1\right)}\)= 5
<=> (x - 1)(x + 1) = 1
<=> x2 -1 = 1
<=> x2 - 2 = 0
<=> (x - \(\sqrt{2}\))(x + \(\sqrt{2}\)) = 0
\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
