Answer:
1. x/x+3 + x+2/x+5
2. x+4/x²+5x+6 • x+3/x²−16
3. 2/x²−9 − 3x/x²−5x+6
4. x+4/x²−5x+6 ÷ x²−16/x+3
Step-by-step explanation:
1. x/x+3 + x+2/x+5
Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)
[Mathematical symbol unavailable]
Therefore, to prove closure, The value of x≠-3 and x≠-5 because if their values are -3 and -5 then the denominator will be zero.
2. x+4/x²+5x+6 • x+3/x²−16
Factors of x²-16 = (x)² -(4)² = (x-4)(x+4)
Factors of x²+5x+6 = x²+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)
Putting factors
[Mathematical symbol unavailable]
Therefore, to prove closure, the value of x≠-2 and x≠4 because if their values are -2 and 4 then the denominator will be zero.
3. 2/x²−9 − 3x/x²−5x+6
Factors of x²-9 = (x)²-(3)² = (x-3)(x+3)
Factors of x²-5x+6 = x²-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)
Putting factors
Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)
[Mathematical symbol unavailable]
Therefore, to prove closure, the value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.
4. x+4/x²−5x+6 ÷ x²−16/x+3
Factors of x²-5x+6 = x²-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)
Factors of x²-16 = (x)² -(4)² = (x-4)(x+4)
Converting ÷ sign into multiplication we will take reciprocal of the second term
[Mathematical symbol unavailable]
Therefore, to prove closure, the value of x≠2 and x≠4 because if their values are 2 and 4 then the denominator will be zero.
