Page 13 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

2
x x + x +1 1 1 1 1 1
2
Solution: = a ⇒ = ⇒ x + = − 1 ⇒ x + =( − 1) − 2.
2
2
x + x +1 x a x a x 2 a
4
2
1 − 2a
1
x +
4 x +1
2
1
Hence, x + x +1 2 2 1 +1 = ( −1) −1= (1 − a) 2 2− a 2 2 = 1 − 2a ⇒ x 2 x 2 = =
(1 − a) − a
1
2
=
= x +
2
+1 = ( −1) −1=
= x +
4 x + x
4
2
x 2 x 2 x 2 x 2 a a a a 2 2 a a 2 2 ⇒ x + x +1 +1
2
a 2 a 2
1 − 2a
1 − 2a .
4
5
3
1.27 A nonzero real number a satisfies a =3a − 1, then find the value of 2a − 5a +2a − 8a 2 .
2
a +1
2
3a
3
2
4
3
2
5
Solution: a =3a − 1 ⇒ a − 3a +1 = 0, and since =1, we have 2a − 5a +2a − 8a 2 (a − 3a + 1)(2a + a +3a) − 3a 3a
2
2
a +1 a +1 = a +1 = − a +1 = −1
2
2
2
2
5
2
3
3
2
4
2a − 5a +2a − 8a 2 (a − 3a + 1)(2a + a +3a) − 3a 3a
= = − = −1 .
2
2
2
a +1 a +1 a +1
y
y + z z + x x + y + z z + x x + y
1.28 Given = = = m and xyz =0, show = m = 2 when
=
=
ay
ay + bz az + bx ax + by + bz az + bx ax + by a+b
a + b =0.
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