Đáp án:
c) `S \ = \ { \ 0 \ ; \ 1/2 \ }`
d) `S \ = \ { \ 2016 \ }`
Giải thích các bước giải:
c)
`text{ĐKXĐ :}` `x ne 1 ; -1`
`(x-3)/(x+1)+(x+2)/(x-1)=5/(x^2-1)`
`<=> ((x-3)(x-1)+(x+2)(x+1))/((x+1)(x-1))=5/((x+1)(x-1))`
`=> (x-3)(x-1)+(x+2)(x+1)=5`
`<=> x^2-4x+3+x^2+3x+2=5`
`<=> 2x^2-x+5=5`
`<=> 2x^2-x=0`
`<=> x(2x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\2x-1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0 \ \ \rm (tm)\\x=\dfrac{1}{2} \ \ \rm (tm)\end{array} \right.\)
Vậy `S \ = \ { \ 0 \ ; \ 1/2 \ }`
$\\$
d)
`(3-x)/2013-1=(2-x)/2014-x/2016`
`<=> (2-x)/2014 - x/2016 - (3-x)/2013 + 1=0`
`<=> ((2-x)/2014+1) + ((-x)/2016-1) + (-(3-x)/2013+1)=0`
`<=> (2016-x)/2014+(2016-x)/2016-(2016-x)/2013=0`
`<=> (2016-x) . (1/2014+1/2016-1/2013)=0`
Mà `1/2014+1/2016-1/2013 ne 0`
`<=> 2016-x=0`
`<=> x=2016`
Vậy `S \ = \ { \ 2016 \ }`
