Understanding Engineering Mathematics

(やまだぃちぅ) #1
(vi)(− 6 )^2

(

3
2

) 3
=(− 1 )^262 (− 1 )^3

33
23

=−

6233
23
on using(− 1 )^2 =1,(− 1 )^3 =− 1

=−

( 2 × 3 )^233
23

=−

223233
23

=−

35
2
(vii) If it helps, just think ofaandbas given numbers:

(−ab^2 )^3
a^2 b

=

(− 1 )^3 a^3 b^6
a^2 b
=−ab^5

(viii) 2^2

( 1
2

)− 3
= 22 ( 2 −^1 )−^3

= 2223 = 25
The steps to watch out for in such problems are the handling of the
minus signs and brackets, and dealing with the negative powers and recip-
rocals.
Don’t forget thatan,n≥0, is not defined fora=0.
B. We can get a long way simply by using



ab=


a


b

(i)


50 =


25 × 2 =


52


2 = 5


2
(ii)


72 −


8 =


36 × 2 −


4 × 2 = 6


2 − 2


2 = 4


2
(iii)(


27 )^3 =( 3


3 )^3 = 33 (


3 )^3 = 333


3 = 34


3 = 81


3

(iv)

(√
2


3
4

) 2
=

(√
3
2


2

) 2
=

3
4 × 2

=

3
8

(v)


3


7

84

=


21

4 × 21

=


21
2


21

=

1
2

(vi) To simplify


3 + 2


2

3 −


2

werationaliseit by removing all surds
from the denominator. To do this we use the algebraic identity:

(a−b)(a+b)≡a^2 −b^2

(see Section 2.2.1) and the removal of surds by squaring. For a√
3 −


2 on the bottom we multiply top and bottom by


3 +


2,
using:

(


3 −


2 )(


3 +


2 )=(


3 )^2 −(


2 )^2 = 3 − 2 = 1
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