Page 11 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

3
1 1 1 3 x
y
1.16 x, y are positive real numbers and − − =0, what is the value of + ?
x y x + y x y

1 1 1 y − x 1 y x y x y x 2 y x √
Solution: − − =0 ⇒ = ⇒ − =1 , thus + = +4 = 5
x y x + y xy x + y x y x y x − y x y

3 2 2
y
x 3

2 3 y x y 2 y x x 2 y x x 2 y x
2
y x y x y x √ . Therefore, y y 3 + x = y + x y − y x + x = + y x + y x − 3 y = x

+ = − +4 = 5 x + = x + 2 x y + 2 = y + y + x y 3 =
x y x y x y y y x − y 2 x x −
2
√ x y √ x y x x y y x y x y x y
√

3 2 √ 5(5 − 3) = 2 5
y x y x y y x x y x y x y x 5(5 − 3) = 2 5 .
3 2 2
+ = + − + = + + − 3 =
x y x y x 2 x y y 2 x y x y x y
√ √
5(5 − 3) = 2 5 1 1 1
+
1.17 Let x,y,z are distinct real numbers, and x,y,z = y + z = z + , show x y z =1.
2 2 2
x
y
y−z
Proof: The conditions imply that x,y,z = x−y , xz = z−x , xy = x−y . Multiply them together
x,y,z
y−z
z−x
to obtain x y z =1.
2 2 2
1.18 Given 2x +6y ≤ 15, x ≥ 0, y ≥ 0, find the maximum value of 4x +3y .
5 1 15 15 5 1
Solution: 2x+6y ≤ 15 ⇒ y ≤ − x ⇒ 4x+3y ≤ 4x+ −x =3x+ ⇒ − x ≥ y ≥ 0,
2 3 2 2 2 3
15 15
thus 4x +3y ≤ 3 × + = 30. The maximum value is 30.
2 2
1.19 Given x + y =8, xy = z + 16, find the value of 3x +2y + z .
2
Solution 1: Let x = 4+ t, y =4 − t , substitute into xy = z + 16: 16 − t = z + 16, which
2
2
2
leads to t = z =0, then x = y =4, thus 3x +2y + z = 12 + 8 + 0 = 20.
2
2
2
2
2
2
Solution 2: Treat x, y as two roots of the equation u − 8u + z + 16 = 0. Δ=64 − 4z − 64 ≥ 0 ⇒ 4z ≤ 0 ⇒ z =0 ⇒ u − 8u + 16 = 0 ⇒ (u − 4) =0 ⇒
2
2
2
2
Δ=64 − 4z − 64 ≥
u 1 = u 2 =4 0 ⇒ 4z ≤ 0 ⇒ z =0 ⇒ u − 8u + 16 = 0 ⇒ (u − 4) =0 ⇒
Δ=64 − 4z − 64 ≥ 0 ⇒ 4z ≤ 0 ⇒ z =0 ⇒ u − 8u + 16 = 0 ⇒ (u − 4) =0 ⇒ u 1 = u 2 =4, i.e. x = y =4.
2
2
2
2

u 1 = u 2 =4
1
1
1
1
1
1
1.20 Given x + y + z =0, find the value of b, x( + )+ y( + )+ z( + ) .
a,
y
z
x
z
x
y
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Solution: b, x( + )+y( + )+z( + )= x( + + )−1+y( + + )−1+z( + + )−1=
1
1
1
1
1
1
1
1
1
1
1
1
a, 1
1
1
( + y )+y( + )+z( + )= x( + + )−1+y( + x+ y)− z1+z( + + )−1=
x
z
z
y
z
y
x
x
z
x
y
y 1 z 1 x 1 z x y x y z x y z x y z
x + )(x + y + z) − 3= 0 − 3= −3.
1
( + 1 ( + 1 y + )(x + y + z) − 3= 0 − 3= −3
z
x y z
be the sum of all digits in n , for instance,
1.21 For a natural number n , let t n
t 2009 = 2 +0+0+9 = 11, evaluate t 1 + t 2 + ··· + t 2009 .
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