Page 22 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

2
2
2
2
2
Proof: N = b−2a+1 = −a +b+a −2a+1 = −(a −b)+(a−1) 2 ⇒ (N + 1)(a − b) =(a − 1) .
2
2
2
a −b a −b a −b
2
Choose N +1 = a − 1, then a − b = a − 1. Thus a = N +2,
b = a − a + 1 = (N + 2) − (N +2)+ 1 = N + 4n + 4 − N − 2 + 1 = N + 3N + 3.
2
2
2
2
Since N is a positive integer, then a, b are are positive integers as well.
√ √
2 4 1+ x 4
1.60 Given x =0, find the maximum value of .
1+ x + x −
x
√ 2 4 √ 4
Solution: 1+ x + x − 1+ x = √ x √
x 1+ x + x + 1+ x 4
4
2
x x
= = whose maximum
1
1
1 2
2
1 2
2
|x|( x + x 2 +1+ x + x 2 ) |x|( (x − ) +3+ (x − ) + 2)
x
x
1 √ √ 1
value is √ √ = 3 − 2 when x, y = x > 0.
3+ 2
1 b + c
2
1.61 Given (b − c) =(a − b)(c − a), a =0, evaluate .
4 a

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