數學問題一問..(thx)萬分感激!!!!!

1)2x^2 - 20x + 32 + 5√(x^2 -10x+25) = 0
2x^2 - 20x + 50 - 18 + 5√(x^2 -10x+25) = 0
2(x^2 -10x+25) + 5√(x^2 -10x+25) - 18 = 0
Let √(x^2 -10x+25) be y :
2y^2 + 5y - 18 = 0
(2y + 9)(y - 2) = 0
√(x^2 -10x+25) = - 9/2 (rejected)
or √(x^2 -10x+25) = 2
x^2 -10x+25 = 4
x^2 - 10x + 21 = 0
(x - 3)(x - 7) = 0
x = 3 or x = 7
2)x^2 -6x-√(2x^2 -12x-5)=20
2x^2 - 12x - 2√(2x^2 -12x-5) = 40
(2x^2 - 12x - 5) - 2√(2x^2 -12x-5) = 40 - 5
Let y be √(2x^2 -12x-5) :
y^2 - 2y - 35 = 0
(y - 7)(y + 5) = 0
√(2x^2 -12x-5) = 7 or - 5(rejected)開方根無負數
2x^2 - 12x - 5 = 49
x^2 - 6x - 27 = 0
(x - 9)(x + 3) = 0
x = 9 or x = - 3




2010-02-04 22:23:21 補充：
001 的答案 1) x=1/2 or 19/2 是增根,不是原方程的解,經平方後所得的解
有可能是假答案,所以要代入試驗。

2)也是。

1.) 個秘缺係前面抽2,
題目=0
->2(x^2-10x+16)+5*sqrt(x^2-10x+25)=0
2(x^2-10x+25)+5*sqrt(x^2-10x+25)-18=0
( 2*sqrt(x^2-10x+25)+9)(sqrt(x^2-10x+25)*--2)=0
sqrt(x^2-10+25)=-9/2 or sqrt(x^2-10x+25)=2
taking the square of both sides, we have,
(x-5)^2=45/4 or (x-5)^2=4
x-5=-9/2 or x-5=9/2 or x-5=-2 or x-5=2
x=1/2 or 19/2 or x=3 or x=7

2. x^2-6x-sqrt(2x^2-12x-5)=20
個秘袂係調轉左右，Sqrt既放一邊，唔係既另一邊,
X^2-6X-20=Sqrt(2x^2-12x-5)
becoz 2x^2-12x-5 cannot be factorized,
we taking square of both sides, we have
(x^2-6x-20)^2=2x^2-12x-5, 好講到呢到其實咁做係錯gei,

2. x^2-6x-sqrt(2x^2-12x-5)=20
-> 1/2(2x^2-12x-5)-sqrt(2x^2-12x-5)=20-5/2
(2x^2-12x-5)-2sqrt(2x^2-12x-5)=35
let sqrt(2x^2-12x-5)=y
y^2-2y=35
y^2-2y-35=0
(y-7)(y+5)=0
y=7 or y=-5, so 2x^2-12x-5=49 or 2x^2-12x-5=25
2x^2-12x-54=0 -(1) or 2x^2-12x-30=0 - (2).
x^2-6x-27=0 or x^2-6x-15=0
(x-9)(x+3)=0 or x=(6+/-sqrt(36+60) )/2
x=9 or x=-3 or x=6+/-4*sqrt(6)/2=2+/-2sqrt(3)



2010-02-05 21:04:42 補充：
哈哈，我錯左，開方根係冇負數既，
所以我輸左la.