mba数学真题3.pdf

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1
   ac− bc + a2 − ab = c( a − b ) + a ( a − b ) = ( a − b )( c + a )
ǎ a− b =3, b + c = − 5 ܖ
a+ c = − 2 ύ =3 × ( − 2) = − 6
Ϥ
స  C
1 1 1
2
    a= x +20, b = x + 19, c = x + 21
20 20 20
 a− b =1, b − c = − 2, a − c = − 1 ܖ
1
a2+ b 2 + c 2 − ab − bc − ac =[( a − b ) 2 + ( b − c ) 2 + ( a − c ) 2 ]
2
1
=(1 + 4 + 1)
2
= 3
Ϥ
స  B
3
    a+ b + c = 0 γ c= − a − b ܖ
a3+ a 2 c − abc + b 2 c + b 3 = a 3 + a 2( − a − b ) − ab ( − a − b ) + b 2 ( − a − b ) + b 3
=a3 − a 3 − a 2 b + a 2 b + ab 2 − ab 2 − b 3 + b 3
= 0
Ϥ
స  E
a − b = 2 + 3
4
    ⇒ a − c = 4 ˣ
b − c = 2 − 3
a2+ b 2 + c 2 − ab − ac − bc =
1 1 1
[2(a 2 + b 2 + c 2 − ab − ac − bc)] = [(a − b) 2 + (b − c) 2 + ( − ac ) 2 ] = [(2 + 3) 2 + (2 − 3) 2 + (4) 2 ] = 15
2 2 2
Ϥ
స  D
5
   a2+ b 2 + c 2 = ab + ac + bc ⇒ a = b = c 
a+2 b + 3 c = 12 ⇒ a = b = c = 2  a+ b2 + c 3 = 2 + 22 + 23 = 14 
Ϥ
స  B
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6
  a 2 + b 2 + c 2 = ab ac ++ bc
 a 2 + b 2 + c 2 − ab − − bcac = 0
⇒ 2a 2 + 2b 2 + 2c 2 − 2ab − 2 − 2bcac = 0