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timurjin [86]
3 years ago
6

WILL GIVE BRAINLIEST

Mathematics
2 answers:
Levart [38]3 years ago
7 0
These steps are easier to follow up too

Anni [7]3 years ago
6 0
Simplify the following:
2 x + 1 + 1/x^2 - (1/x^2 + x - 2)

Put each term in x - 2 + 1/x^2 over the common denominator x^2: x - 2 + 1/x^2 = x^3/x^2 - (2 x^2)/x^2 + 1/x^2:
2 x + 1 + 1/x^2 - x^3/x^2 - (2 x^2)/x^2 + 1/x^2

x^3/x^2 - (2 x^2)/x^2 + 1/x^2 = (x^3 - 2 x^2 + 1)/x^2:
1/x^2 + 2 x + 1 - (x^3 - 2 x^2 + 1)/x^2

The possible rational roots of x^3 - 2 x^2 + 1 are x = ± 1. Of these, x = 1 is a root. This gives x - 1 as all linear factors:
1/x^2 + 2 x + 1 - (((x - 1) (x^3 - 2 x^2 + 1))/(x - 1))/(x^2)

 | | 
x | - | 1 | | x^2 | - | x | - | 1
x^3 | - | 2 x^2 | + | 0 | + | 1
x^3 | - | x^2 | | | | 
 | | -x^2 | + | 0 | | 
 | | -x^2 | + | x | | 
 | | | | -x | + | 1
 | | | | -x | + | 1
 | | | | | | 0:
1/x^2 + 2 x + 1 - (x^2 - x - 1 (x - 1))/x^2

Put each term in x^(-2) + 2 x + 1 - ((x - 1) (x^2 - x - 1))/x^2 over the common denominator x^2: x^(-2) + 2 x + 1 - ((x - 1) (x^2 - x - 1))/x^2 = 1/x^2 + (2 x^3)/x^2 + x^2/x^2 - ((x - 1) (x^2 - x - 1))/x^2:1/x^2 + (2 x^3)/x^2 + x^2/x^2 - ((x - 1) (x^2 - x - 1))/x^2
1/x^2 + (2 x^3)/x^2 + x^2/x^2 - ((x - 1) (x^2 - x - 1))/x^2 = (1 + 2 x^3 + x^2 - (x - 1) (x^2 - x - 1))/x^2:(1 + 2 x^3 + x^2 - (x - 1) (x^2 - x - 1))/x^2

 | | | | x | - | 1
 | | x^2 | - | x | - | 1
 | | | | -x | + | 1
 | | -x^2 | + | x | + | 0
x^3 | - | x^2 | + | 0 | + | 0
x^3 | - | 2 x^2 | + | 0 | + | 1:
(-x^3 - 2 x^2 + 1 + 2 x^3 + x^2 + 1)/x^2


-(x^3 - 2 x^2 + 1) = -x^3 + 2 x^2 - 1:
(-x^3 + 2 x^2 - 1 + 2 x^3 + x^2 + 1)/x^2


Grouping like terms, 2 x^3 - x^3 + 2 x^2 + x^2 - 1 + 1 = (2 x^3 - x^3) + (x^2 + 2 x^2) + (1 - 1):
((2 x^3 - x^3) + (x^2 + 2 x^2) + (1 - 1))/x^2


2 x^3 - x^3 = x^3:
(x^3 + (x^2 + 2 x^2) + (1 - 1))/x^2


x^2 + 2 x^2 = 3 x^2:
(x^3 + 3 x^2 + (1 - 1))/x^2


1 - 1 = 0:
(x^3 + 3 x^2)/x^2


Factor x^2 out of x^3 + 3 x^2:
(x^2 (x + 3))/x^2


(x^2 (x + 3))/x^2 = x^2/x^2×(x + 3) = x + 3:
Answer:  x + 3
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