Simplify the following :
(1) [(x2 - 2x)/(x + 2)] ⋅ [(3x + 6)/(x - 2)]
(2) [(x2 - 81)/(x2 - 4)] ⋅ [(x2 + 6x + 8)/(x2 - 5x - 36)]
(3) [(x2 - 3x - 10)/(x2 - x - 20)] ⋅ [(x2 - 2 x + 4)/(x3 + 8)]
(4) [(x2 - 16)/(x2 - 3x + 2)] ⋅ [(x2 - 4)/(x3 + 64)] ⋅
[(x2 - 4x + 16)/(x2 - 2x - 8)]
(5) [(3x2 + 2x - 1)/(x2 - x - 2)] [(2x2 - 3 x - 2)/(3x2 + 5x - 2)]
(6) [(2x - 1)/(x2 + 2x + 4)] ⋅[(x4 -8 x)/(2x2 + 5x - 3)] ⋅
[(x + 3)/(x2 - 2x)]
(7) [(a + b)/(a - b)] [(a3 - b3)/(a3 + b3)]
(8) [(x2 - 9y2)/(3x - 3y)] ⋅ [(x2 - y2)/(x2 + 4xy + 3y2)]
(9) [(x2 - 4x - 12)/(x2 - 3x - 18)] ⋅ [(x2 - 2x - 3)/(x2 + 3x + 2)]
(10) [(x2 - 3x - 10)/(x2 - x - 20)]⋅[(x2 - 4x + 16)/(x3 + 64)]
(11) [(x2 - 16)/(x - 2)] [(x2 - 4)/(x3 + 64)]
(12) [(x + 7)/(x2 + 14x + 49)] [(x2 + 8x + 7)/(x + 1)]
Problem 1 :
[(x2 - 2x)/(x + 2)] ⋅ [(3x + 6)/(x - 2)]
Solution :
Let f(x) = [(x2 - 2x)/(x + 2)] ⋅ [(3 x + 6)/(x - 2)]
f(x) = [(x2 - 2x)/(x + 2)] ⋅ [(3x + 6)/(x - 2)]
f(x) = [x(x - 2)/(x + 2)] ⋅ [3(x + 2)/(x - 2)]
f(x) = 3x
So, the value of f(x) is 3x.
Problem 2 :
[(x2 - 81)/(x2 - 4)] ⋅ [(x2 + 6x + 8)/(x2 - 5x - 36)]
Solution :
Let f(x) = [(x2 - 81)/(x2 - 4)] ⋅ [(x2 + 6x + 8)/(x2 - 5x - 36)]
x2 - 81 = x2 - 92 ==> (x + 9)(x - 9)
x2 - 4 = x2- 22 ==> (x + 2)(x - 2)
x2 + 6x + 8 = (x + 2)(x + 4)
x2 - 5x - 36 = (x - 9)(x + 4)
f(x) = [(x + 9)(x - 9)/(x + 2)(x - 2)] ⋅ [(x + 2)(x + 4)/(x - 9)(x + 4)]
By simplifying (x + 9)/(x - 2)
So, the value of f(x) is (x + 9)/(x - 2).
Problem 3 :
[(x2 - 3x - 10)/(x2 - x - 20)] ⋅ [(x2 - 2x + 4)/(x3 + 8)]
Solution :
Let f(x) = [(x2 - 3x - 10)/(x2 - x - 20)] ⋅ [(x2- 2 x + 4)/(x3 + 8)]
x2 - 3x - 10 = (x - 5)(x + 2)
x2 -x - 20 = (x - 5)(x + 4)
a3+b3 = (a+b)(a2-ab+b2)
x3+23 = (x+2)(x2-2x+4)
By applying the factors in f(x), we get
= [(x - 5)(x + 2)/(x - 5)(x + 4)] ⋅ [(x2 - 2x + 4)/(x + 2)(x2-2x + 4)]
= 1/(x + 4)
So, the value of f(x) is 1/(x + 4)
Problem 4 :
[(x2 - 16)/(x2 - 3x + 2)] ⋅ [(x2 - 4)/(x3 + 64)] ⋅
[(x2 - 4x + 16)/(x2 - 2x - 8)]
Solution :
Let f(x) = [(x2 - 16)/(x2 - 3x + 2)] ⋅ [(x2 - 4)/(x3 + 64)] ⋅
[(x2 - 4x + 16)/(x2 - 2x - 8)]e
x2-16 = x2 - 42 ==> (x + 4)(x - 4)
x2 - 3x + 2 = (x - 1)(x - 2)
x2 - 4 = x2 - 22 ==> (x + 2)(x - 2)
x3+64 = x3+43 ==> (x+4)(x2-4x+16)
x2-2x-8 = (x-4)(x+2)
= [(x+4)(x-4)/(x-1)(x-2)]⋅[(x+2)(x-2)/(x+4)(x2-4x+16)]
⋅[(x2-4x+16)/(x-4)(x+2)]
f(x) = 1/(x-1)
So, the value of f(x) is 1/(x-1).
Problem 5 :
[(3x2+2x-1)/(x2-x-2)] [(2x2-3x-2)/(3x2+5x-2)]
Solution :
Let f(x) = [(3x2+2x-1)/(x2-x-2)]⋅
[(2x2-3x-2)/(3x2+5x-2)]
(3x2+2x-1) = (3x-1) (x+1)
(x2-x-2) = (x-2) (x+1)
(2x2-3x-2) = (2x+1) (x-2)
(3x2+5x-2) = (2x-1) (x+2)
By applying the factors in f(x), we get
= [(3x-1)(x+1)/(x-2) (x+1)]⋅[(2x+1) (x-2)/(2x-1) (x+2)]
= (2x+1)/(x+2)
So, the value of f(x) is (2x+1)/(x+2).
Problem 6 :
[(2x-1)/(x2+2x+4)] ⋅[(x4-8x)/(2x2+5x-3)] ⋅
[(x+3)/(x2-2x)]
Solution :
Let f(x) = [(2x-1)/(x2+2x+4)] ⋅[(x4-8x)/(2x2+5x-3)] ⋅
[(x+3)/(x2-2x)]
x4-8x = x(x3-23)
x4-8x = x(x-2)(x2+2x+4)
2x2+5x-3 = (2x-1)(x+3)
x2-2x = x(x-2)
By applying the factors in f(x), we get
= [(2x-1)/(x2+2x+4)]⋅[x(x-2)(x2+2x+4)/(2x-1)(x+3)] ⋅
[(x+3)/x(x-2)]
= 1
So, the value of f(x) is 1.
Problem 7 :
[(a+b)/(a-b)] [(a3-b3)/(a3+b3)]
Solution :
Let f(x) = [(a+b)/(a-b)] [(a3-b3)/(a3+b3)]
= [(a+b)/(a-b)]⋅[(a-b)(a2+ab+b2)/(a+b) (a2-ab+b2)]
= (a2+ab+b2)/(a2-ab+b2)
So, the value of f(x) is (a2+ab+b2)/(a2-ab+b2).
Problem 8 :
[(x2-9y2)/(3x-3y)] ⋅ [(x2-y2)/(x2+4xy+3y2)]
Solution :
Let f(x) = [(x2-9y2)/(3x-3y)] ⋅ [(x2-y2)/(x2+4xy+3y2)]
x2-9y2 = x2-(3y)2
x2-9y2 = (x+3y)(x-3y)
3x-3y = 3(x-y)
x2-y2 = (x+y)(x-y)
x2+4xy+3y2 = (x+3y)(x+y)
By applying the factors in f(x), we get
= [(x+3y)(x-3y)/3(x-y)]⋅[(x+y)(x-y)/(x+3y)(x+y)]
By simplifying, we get
= (x-3y)/3
So, the value of f(x) is (x-3y)/3.
Problem 9 :
[(x2-4x-12)/(x2-3x-18)] ⋅ [(x2-2x-3)/(x2+3x+2)]
Solution :
Let f(x) = [(x2-4x-12)/(x2-3x-18)]
⋅ [(x2-2x-3)/(x2+3x+2)]
x2-4x-12 = (x-6)(x+2)
x2-3x-18 = (x-6)(x+3)
x2-2x-3 = (x-3)(x+1)
x2+3x+2 = (x+1)(x+2)
f(x) = [(x-6)(x+2)/(x-6)(x+3)]⋅[(x-3)(x+1)/(x+1)(x+2)]
f(x) = (x-3)/(x+3)
So, the value of f(x) is (x-3)/(x+3).
Problem 10 :
[(x2-3x-10)/(x2-x-20)]⋅[(x2-4x+16)/(x3+64)]
Solution :
Let f(x) = [(x2-3x-10)/(x2-x-20)]⋅[(x2-4x+16)/(x3+64)]
x2-3x-10 = (x-5)(x+2)
x2-x-20 = (x-5)(x+4)
x3+43 = (x+4)(x2-4x+16)
By applying the factors in f(x), we get
f(x) = [(x - 5)(x + 2)/(x - 5)(x + 4)]⋅[(x2 - 4x + 16)/(x + 4)(x2 - 4x + 16)]
f(x) = (x + 2)/(x + 4)2
So, the value of f(x) is (x+2)/(x+4)2.
Problem 11 :
[(x2-16)/(x-2)] [(x2-4)/(x3+64)]
Solution :
Let f(x) = [(x2 - 16)/(x - 2)] [(x2 - 4)/(x3 + 64)]
x2-16 = x2 -4 2 ==> (x + 4)(x - 4)
x2-4 = x2 - 22 ==> (x + 2)(x - 2)
x3+64 = x3 + 43 ==> (x + 4)(x2 - 4x + 16)
f(x) = [(x + 4)(x - 4)/(x - 2)] [(x + 2)(x - 2)/(x + 4)(x2-4x + 16)]
f(x) = (x - 4)(x - 2)/(x2 - 4x + 16)
So, the value of f(x) is (x - 4)(x - 2)/(x2 - 4x + 16).
Problem 12 :
[(x + 7)/(x2 + 14x + 49)] [(x2 + 8x + 7)/(x + 1)]
Solution :
Let f(x) = [(x + 7)/(x2 + 14x + 49)] [(x2 + 8x + 7)/(x + 1)]
x2 + 14x + 49 = (x + 7)(x + 7)
x2 + 8x + 7 = (x + 1)(x + 7)
By applying the factors in f(x), we get
f(x) = [(x + 7)/(x + 7)(x + 7)] [(x + 1)(x + 7)/(x + 1)]
f(x) = 1
So, the value of f(x) is 1.
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