Page 36 - Elementary Algebra Exercise Book I
P. 36
ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers
2
2
2
1.96 The real numbers x,y,z satisfy x + y + z =3 (i), x + y + z = 29 (ii),
there are 2000 integers satisfying the inequality. If x< 0, then the inequality becomes
3 (2000 − x)+ (−x) ≤ 9999 ⇔−3999.5 ≤ x< 0.
4There are additionally 3999 integers
3
x + y + z = 45 (iii). Evaluate xyz and x + y + z .
3
4
4
satisfying the inequality. Hence, totally there are 4000 + 2000 + 3999 = 9999 integers
2 inequality.
satisfying the
Solution: (i) -(ii): xy + yz + zx = −10. Since
3
3
2
3
3
3
3
2
2
2
x + y + z − 3xyz = x +3x y +3xy + y + z − 3x y − 3xy − 3xyz =(x +
3
3
2
3
3
2
2
2
3
3
3 ⋆⋆⋆
3 x + y + z − 3xyz = x +3x y +3xy + y + z − 3x y − 3xy − 3xyz =(x +
2
2
2
3
2 2
3 real numbers x, y, z satisfy x + y + z = 3 (i), x + y + z = 29 (ii),
The
1.96
y) + z − 3xy(x + y + z) =(x + y + z)[(x + y) − (x + y)z + z ] − 3xy(x + y + z)=
3
2
2
3
2
3
3
2
x + y + z − 3xyz = x +3x y +3xy + y + z − 3x y − 3xy − 3xyz =(x +
2 3xy(x + y + z) =(x + y + z)[(x + y) − (x + y)z + z ] − 3xy(x + y + z)=
3
3
3
3
3
y) + z − z = 45
x + y +
2(iii). Evaluate xyz and x + y + z .
2
(x + y + z)(x + y + z − xy − yz − zx) 4 4 4 2 2 2 2
3
3
y) + z − 3xy(x + y + z) =(x + y + z)[(x + y) − (x + y)z + z ] − 3xy(x + y + z)=
2
2
2
(x + y + z)(x + y + z − xy − yz − zx)
(x + y + z)(x + y + z − xy − yz − zx) , then 45 − 3xyz = 3(29 + 10) ⇒ xyz = −24.
2
2
2
2
3
3
2
3
3
Solution: (i) -(ii): xy + yz + zx = −10. Since x + y + z − 3xyz = x +3x y +
2
3
3
2 3 3 2 2 3xy − 3xyz =(x + y) + z − 3xy(x + y + z) =(x + y + z)[(x +
2 2
2 2
2 2
Since (xy + yz + zx) = 100 ⇒ x y + y z + z x +2xyz(x + y + z) = 100, then
3xy + y + z − 3x y −
2
2
2
2
2
y) − (x + y)z + z ] − 3xy(x + y + z)= (x + y + z)(x + y + z − xy − yz − zx), then
2 2
2 2
2 2
2
2 2
2 2
x y + y z + z x = 100 − 2 · (−24) · 3 = 244. yz + zx) = 100 ⇒ x y + y z +
45 − 3xyz = 3(29 + 10) ⇒ xyz = −24. Since (xy +
2 2
2 2
2 2
2 2
z x +2xyz(x + y + z) = 100, then x y + y z + z x = 100 − 2 · (−24) · 3 = 244.
2 2
2
2 2
2 2
2
2 2
2
4
Hence, x + y + z =(x + y + z ) − 2(x y + y z + z x ) = 29 − 2 × 244 = 353.
4
4
4
2
2
2 2
2 2
4
2 2
2
4
2 2
Hence, x + y + z =(x + y + z ) − 2(x y + y z + z x ) = 29 − 2 × 244 = 353.
1 1
+ 1 1 1 1 1 1
= 1+
2 + ··· +
2 +
1.97 ⋆ Let S = 2 + 2009 , find [S].
1.97 Let S min 1+ S max 3 + · ·· + 2 , find [S].
2 2 3 2 2009 2
1
1
1
1
1
1
1
1 1
1
+ ··· +
+
Solution: 1 <S = 1+ 2 1 2 2 1 2 + ··· +·+ 1 1 2 2 < 1+ < 1+ + + 1 1 1 + +·· ·+ + ··· + 1 2008 × 2009 =
<
+ +
+ ··· +
+
=
2 1+
Solution: 1 <S
=
1 <S = 1+ = 1+
2 × 3
3
1 × 2
2009
2 ··
2
2 × 3 3
2009
2
2008 × 2009
2008 × 2009
2 ×
1 ×
1
2008
1
1 1 1 1 1 1 1 2 1 2 3 1 3 1 1 2009 1 1 1 × 2 2 2008 =1 . Hence, [S]=1.
1 1
1
2008
=1
+ +
1+1
−
+ ··· +
=2 −
−
1+1 − − 1+1 − − − 2 + + ··· + + ··· + − 2008 − =2 − =2 − =1 . Hence, [S] = 1.
2009
2
2009
3
2009
2 2 2 2 3 3 2008 2009 2009 2009
2009
2009
2009
2008
1 1
+ 1 1 1 1 1 1
1.98 ⋆⋆⋆ = 1+ √ + √ + ··· + √ , find [S].
Let S =
1.98 Let S min 1+ √ + √ + ·· · + √ 994009 , find [S].
S max
2
3
2 3 994009
1 1 1
Solution: Let k be a positive integer, we have √ √ < √ < √ √ ⇔
k +1+ k 2 k k + k − 1
√ √ 1 √ √ √ 1 √ √
k +1 − k< √ < k − k − 1. Thus we have 2 − 1 < √ < 1, 3 − 2 <
2 k 2 1
1 √ √ √ 1 √ √
√ < 2 − 1, ·· · , 994010 − 994009 < √ < 994009 − 994008. Add
2 2 2 994009
√ 1 1 1 1 √ 1
them up to get 994010 − 1 < (1 + √ + √ + ·· · + √ ) < 994009 − ⇒
2 2 3 994009 2
1 1 √
997 − 1 < S< 997 − ⇒ 1992 < 2 994010 − 2 <S < 1993 ⇒ [S] = 1992.
2 2
1.99 Given x,y,z,a,b,c are distinct rewal numbers, and
1 1 1 1
+ + = ,
x + a y + a z + a a
1 1 1 1
+ + = ,
x + b y + b z + b b
1 1 1 1
+ + = ,
x + c y + c z + c c
1 1 1
Evaluate + + .
a b c
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