Page 60 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 58

2
( m + n x + m – 2n + p x + m – 2p )
(
)
)
(
= ( )
2
x – 2 x + x + 1
( )

 m + n = 1 ⇒ n = 1 – m (1)







ផ។ទឹម គុណ ើងបាន  m – 2n + p = 6 (2)


 m – 5

 m – 2p = 5 ⇒ p = (3)


2
m – 5
យក (1); (3) ជួសក៖ន ង (2) ើងបាន m – 2(1 – m) + = 6
2
m – 5
m – 2 + 2m + = 6
2
2m – 4 + 4m + m – 5
2 = 6
7m – 9 = 12
21
⇒7m = 21 ⇒ m = = 3
7
m = 3 បាន n = 1 – m = 1 – 3 = –2
m – 5 3 – 5
p = = = –1
2 2
ដូច ះ m = 3 ; n = –2 ; p = –1
1 x + 6x + 5
2
∫
២. គណនាអាំង ល I = ( )dx
2
0 (x – 2) x + x + 1
2
x + 6x + 5 3 –2x – 1
យ ( ) = +
2
2
(x – 2) x + x + 1 x – 2 x + x + 1
2
1 x + 6x + 5 1 3 –2x – 1
∫ ∫ ( )
បាន I = ( )dx = + dx
2
2
0 (x – 2) x + x + 1 0 x – 2 x + x + 1
′
 ( ) 
∫ 2
1   3(x – 2) ′ x + x + 1   [ 1
2
]
=      – 2    dx = 3 ln |x – 2| – ln x + x + 1


0  (x – 2) x + x + 1  0
3
)
)
(
(
= 3 ln 1 – ln 3 – 3 ln 2 – ln 1 = – ln 3 – ln 2 = – ln 3 + ln 8 = – ln 24
I = – ln 24
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