數學題求解答!急!!!!!!

2007-12-26 7:15 am
已知:

x²+y²-6x+2y+10=0

求:

(3x²-2xy- y²)/(5x²-7xy+2y²)=?

回答 (2)

2007-12-30 2:00 am
✔ 最佳答案
已知:

x^2 + y^2 - 6x + 2y + 10 = 0

求:

( 3x^2 - 2xy - y^2 ) / ( 5x^2 - 7xy + 2y^2 ) = ?

Sol:

x^2 + y^2 - 6x + 2y + 10 = 0

x^2 - 6x + 9 + y^2 + 2x + 1 = 0

( x - 3 )^2 + ( y + 1 )^2 = 0

x = 3, y = - 1

( because the square of any real number always ≧ 0 )

( 3x^2 - 2xy - y^2 ) / ( 5x^2 - 7xy + 2y^2 )

= [ 3 * 3^2 - 2 * 3 * ( - 1 ) - ( - 1 )^2 ] / [ 5 * 3^2 - 7 * 3 * ( - 1 ) + 2 * ( - 1 )^2 ]

= ( 27 + 6 - 1 ) / ( 45 + 21 + 2 )

= 32 / 68

= 8 / 17

Ans: 8 / 17
參考: 數學小頭腦
2007-12-30 8:33 pm
x^2 + y^2 - 6x + 2y + 10 = 0
x = 3, y = - 1
( 3x^2 - 2xy - y^2 ) / ( 5x^2 - 7xy + 2y^2 )
= ( 27 + 6 - 1 ) / ( 45 + 21 + 2 )
= 8 / 17
參考: 自己


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