Page 20 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

a b a b
Proof: = ⇒ = . Since x, y = a+b ,y = b+c ,
b c a + b b + c 2 2
a c a c 2a 2c 2b 2c
we have + = + = + = + =2.
x y (a + b)/2 (b + c)/2 a + b b + c b + c b + c
2012 2012 2012
1.51 Given abc =1, evaluate + + .
1+ a + ab 1+ b + bc 1+ c + ca

Solution: 2012 + 2012 + 2012
1+ a + ab 1+ b + bc 1+ c + ca

1 1 1
= 2012 + +
1+ a + 1 1+ 1 + 1 c 1+ c + ca
c c ac ac 1
ac
= 2012 + + = 2012 .
c + ac +1 ac +1+ c 1+ c + ca

7
6
1.52 Given x + x + x = −1, evaluate x 2000 + x 2001 + ··· + x 2012 .
Solution: x + x + x = −1 ⇒ x (x + 1) + (x + 1) = 0 ⇒ (x + 1)(x + 1) = 0 ⇒ x = −1 since
7
6
6
6
6
x +1 > 0. Hence, x 2000 + x 2001 + ··· + x 2012 = 1+ 1+ ··· +1 + (−1) + (−1) + ··· +(−1) =1.

seven 1 s six (−1) s

1 1 1
1.53 Given a<b< c , determine ± sign of + + .
a − b b − c c − a
Solution: Let a − b = x, b − c = y, c − a = z. Since a<b< c , we have x< 0,y < 0,z > 0.
2
x + y + z = a − b + b − c + c − a =0 ⇒ (x + y + z) =0 ⇒ 2(xy + yz + zx)=
2 x + y +
2z = a − b + b − c + c − a =0 ⇒ (x + y + z) =0 ⇒ 2(xy + yz + zx)=
2
−(x + y + z ) ⇒ xy + yz + zx < 0 2
−(x + y + z ) ⇒ xy + yz + zx < 0. Therefore
2
2
2
1 1 1 1 1 1 yz + zx + xy , that is, 1 1 1 is negative.
+ + = + + = < 0 + +
a − b b − c c − a x y z xyz a − b b − c c − a
1.54 Factor (a + b − 2ab)(a − 2+ b) + (1 − ab) .
2
Solution: Let a + b = x, ab = y , then

2
2
2
(a + b − 2ab)(a − 2 + b) + (1 − ab) =(x − 2y)(x − 2) + (1 − y) = x − 2x − 2xy +
2
2
2
2
2
4y +1 − 2y + y = x − 2x(y +1) +(y + 1) =(x − y − 1) =(a + b − ab − 1) =
2
2
[(a − 1) − b(a − 1)] =(a − 1) (b − 1) 2
1.55 The real numbers m, n, p are not all equal, and x = m − np, y = n − pm, z = p − mn .
2
2
2

Show at least one of x,y,z is positive.

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