Page 21 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Proof: 2(x+y +z)=2(m +n +p −mn−np−pm) =(m−n) +(n−p) +(p−m) ≥ 0 .
2
2
2
2
2
2

In addition, since m, n, p are not all equal, then m − n, n − p, p − m are not all zeros.
Thus x + y + z> 0, which shows at least one of x,y,z is positive.
1.56 a, b, c are nonzero real numbers, and a + b + c =0,

1 1 1
evaluate + + .
2
2
2
2
2
2
b + c − a 2 c + a − b 2 a + b − c 2
2
2
Solution: a + b + c =0 ⇒ b + c = −a ⇒ (b + c) = a ⇒ b + c − a = −2bc . Similarly,
2
2
2
2
2
2
2
we can obtain a + b − c = −2ab, c + a − b = −2ca. In addition, −2bc, −2ab, −2ca are
2
2
1 1 1 1 1 1 a + b + c
nonzero. Hence, + + = − − − = − =0.
2
2
2
2
2
2
b + c − a 2 c + a − b 2 a + b − c 2 2bc 2ca 2ab 2abc
2
1.57 Find the minimum value of the fraction 3x +6x +5 .
1 x + x +1
2
2
2
2
2
3x +6x +5 6x + 12x + 10 6(x +2x + 2) − 2 2
Solution: 1 2 = 2 = 2 =6 − 2 which
2 x + x +1 x +2x +2 x +2x +2 (x + 1) +1
has the minimum value 4 when x = −1.
1.58 For real numbers x, y , define the operator x ∗ y = ax + by + cxy where a, b, c
are constants. We know that 1 ∗ 2 =3, 2 ∗ 3= 4, and there is a nonzero real number d
such that x ∗ d = x holds for any real number x. Find the value of d.
Solution: For any real number x , we have x ∗ d = ax + bd + cdx = x , 0 ∗ d = bd =0.
Since d =0, then b =0, thus
1 ∗ 2= a +2b +2c =3
2 ∗ 3 =2a +3b +6c =4
⇒
a +2c =3
2a +6c =4

which results in a =5,c = −1 . In addition, 1 ∗ d = a + bd + cd =1 , and substitute
a =5,b =0,c = −1 into it to obtain d =4.

1.59 Show for any positive integer N , we can find two positive integers a and b
b − 2a +1
such that N = .
a − b
2

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