Page 30 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Solution: M = N implies that one element in M should be 0. The existence of lg(xy)
1
implies that xy =0, thus x, y cannot be 0. Hence, lg(xy)=0 ⇒ xy =1 ⇒ y = .
x

1
Thus M = {x, 1, 0},N = {0, |x|, } . According to M = N again, we have either
x

x = |x|
1
1=
x

or

1
x =
x
1= |x|

However, x =1 violates the uniqueness of every element in a set. Hence, x = −1,y = −1.

1
1
2k+1
Then xy =0 + y 2k+1 = −2 (k =0, 1, 2, ··· ); b, x 2k + y 2k =2 (k =1, 2, ··· ). In the original
a,
summation, the number of 2k +1 terms is one more than the number of 2k terms, therefore
1 1 1 1
3
2
(x + ) + (x + ) +(x + )+ ··· +(x 2001 + )= −2 .
y y 2 y 3 y 2001
√ √
6
1.81 Find the integer part of the number ( 7+ 5) .
√ √ √ √
Solution: Let 7+ 5= x, 7 − 5= y , then
√ √
2 2 2 2
x + y = 2 7, xy = 2 ⇒ x + y = (x + y) − 2xy = (2 7) − 2 × 2 = 24 ⇒ .
2 2
2
2
2
2 2
2
4
2
4
2 2
6
2 3
2 3
6
x + y =(x ) +(y ) =(x + y )(x − x y + y )=(x + y )[(x + y ) − 3x y ]=
2
24[24 − 3 × 4] = 24 × 564 = 13536
√ √ √ √ √ √
6
6
Hence, ( 7+ 5) +( 7 − 5) = 13536. Since 0 < 7 − 5 < 1,
√ √ √ √
6
6
then 13535 < ( 7+ 5) < 13536, which implies that the integer part of ( 7+ 5) is
13535.
1.82 The real numbers x, y satisfy 4x − 5xy +4y =5, let S = x + y , evaluate
2
2
2
2
1 1
+ .
S min S max
Solution 1: Let x = a + b, y = a − b and substitute into 4x − 5xy +4y =5 :
2
2
2
2 5 − 3a
2
2
2
2
2 2
2
2
2
4(a + b) − 5(a + b)(a − b) + 4(a − b) =5 ⇒ 3a + 13b =5 ⇒ b = 5 − 3a 2 ≥ 0 ⇒
4(a + b) − 5(a + b)(a − b) + 4(a − b) =5 ⇒ 3a + 13b =5 ⇒ b =
≥ 0 ⇒
1 1 13 13
5 5 + 2
a + .
2
2
2
2
2
2 2 2 2 =(a + b) +(a − b) =2a +2b =2a + 10−6a = 20 2 10
5 − 3a ≥ 0 ⇒ 0 ≤ a ≤ . Therefore S min
5 − 3a ≥ 0 ⇒ 0 ≤ a ≤ S max 13 13 13
3
1 3 1 1 1
+ + 1 1 13 3 8
min = 10 2 5 max = 10 . Hence, + = + = .
When a =0, S min S max 3 S max
. When a =0 = , S min
13
3
S min S max 10 10 5
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