2.4. Rational Expressions http://www.ck12.org
=
( 1
x+ 1 −1
x+ 2)
÷
( 1
x− 2 +1
x+ 1)
=
[ (x+ 2 )
(x+ 1 )(x+ 2 )−(x+ 1 )
(x+ 1 )(x+ 2 )]
÷
[ (x+ 1 )
(x+ 1 )(x− 2 )−(x− 2 )
(x+ 1 )(x− 2 )]
=
[ 1
(x+ 1 )(x+ 2 )]
÷
[ 3
(x+ 1 )(x− 2 )]
=(x+ 1 )(^1 x+ 2 )·(x+^1 )( 3 x−^2 )= 3 ((xx−+^22 ))- Being able to factor effectively is of paramount importance.
=xx−+^23 −x(^3) − 3 x (^2) + 8 x− 24
2 (x+ 2 )(x^2 − 9 )
=((xx−+^23 ))−x
(^2) (x− 3 )+ 8 (x− 3 )
2 (x+ 2 )(x^2 − 9 )
=((xx−+ 32 ))− (x−^3 )(x
(^2) + 8 )
2 (x+ 2 )(x+ 3 )(x− 3 )
Before subtracting, simplify where possible so you don’t contribute to unnecessarily complicated denominators.
=((xx−+^23 ))− x
(^2) + 8
2 (x+ 2 )(x+ 3 )
The left expression lacks 2(x+ 2 ), so multiply both its numerator and denominator by 2(x+ 2 ).
=^22 ((xx++ 22 )()(xx−+^23 ))− (x
(^2) + 8 )
2 (x+ 2 )(x+ 3 )
=^2 (x
(^2) − 4 )−x (^2) − 8
2 (x+ 2 )(x+ 3 )
= x
(^2) − 16
2 (x+ 2 )(x+ 3 )
- It would be an exercise in futility to attempt to try to compute this expression directly. Instead, notice that the
repeating nature of the expression lends itself to an extremely nice substitution.
Let 2 + −^1 − 1
2 + 2 + −−^11
2 + 2 +− 21 −+ (^1) ...
=x
Notice that the red portion of the fraction is exactly the same as the rest of the fraction and soxmay be substituted
there and solved.