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幾題復雜少少二次方程(20點)
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1.已知f(x)=2(x+m)和g(x)=x+m。若f(8)=6g(4)，求m的值。答案是-22.已知f(x)=(x+2)(x-2)+ax+b。若f(2)=2和f(-2)=4,求a和b的值。答案是a=-(1/2),b=33.設f(x)=3x-1和g(x)=x-2。(a)若2f(a)-3g(a)=5,求a的值。答案是1/3(b)若f(2b)-g(3b)=5,求b的值。答案是4/34.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。(a)求k的值。答案是-2(b)由此，若2x-f(x)=0,求x的值。答案是-1,-25.已知f(x)=x^2-kx。(a)求f(x+2)和f(x-... 顯示更多 1.已知f(x)=2(x+m)和g(x)=x+m。若f(8)=6g(4)，求m的值。答案是-2 2.已知f(x)=(x+2)(x-2)+ax+b。若f(2)=2和f(-2)=4,求a和b的值。答案是a=-(1/2),b=3 3.設f(x)=3x-1和g(x)=x-2。 (a)若2f(a)-3g(a)=5,求a的值。答案是1/3 (b)若f(2b)-g(3b)=5,求b的值。答案是4/3 4.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。 (a)求k的值。答案是-2 (b)由此，若2x-f(x)=0,求x的值。答案是-1,-2 5.已知f(x)=x^2-kx。 (a)求f(x+2)和f(x-2)的值。答案是f(x+2)=x^2+(4-k)x+4-2k,f(x-2)=x^2-(4+k)x+4+2k (b)若f(x+2)-f(x-2)=kx-32,求k的值。答案是8 (c)由此，若f(x+2)=f(x-2)+40,求x的值。答案是9 更新: to N.Y Cafe: 第2題中我不太明白你點樣從 「(2) - (1) -2a+b=4 直式加減」 做到 「(-) 2a+b=2 -4a =2 a= -1/2 b=3」 唔該,你可唔可以從(2) - (1)個到做詳細清楚少少呀,多謝指教!!! 更新 2: 4.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。 (b)由此，若2x-f(x)=0,求x的值。答案是-1,-2 我想問個第3頂不是 - 嗎?為何會有個+出現?因為個問題都是-k^2,自然是-(-2)^2.... 2x- (x+3)(x+2) +(-2)^2 = 0
最佳解答:
1) f(x)=2(x+m)和g(x)=x+m f(8)=6g(4) 左方：f(8)=2(8+m) ←代x係8 =16+2m 右方：g(4)=6(4+m) ←代x係4，同埋係6倍！ =24+6m ∵16+2m=24+6m 16-24=6m-2m -8=4m m= -2 2) f(x)=(x+2)(x-2)+ax+b f(2)=2和f(-2)=4 ∵f(2)=2 (2+2)(2-2)+a(2)+b=2 2a+b=2 ←(1) ∵f(-2)=4 (-2+2)(-2-2)+a(-2)+b=4 -2a+b=4 ←(2) (2) - (1) -2a+b=4 直式加減 -) 2a+b=2 -4a =2 a= -1/2 b=3 如果唔介意，我聽日再答埋你其他問題！ 2007-11-18 19:48:52 補充： 好似聯立方程咁呢！只要將D位對返正你應該可以計到嫁(2) - (1) -2a b=4 直式加減-) 2a b=2 -4a =2a= -1/2 ←代返個a入(2a b=2)或代入( -2a b=4)b=3 ←得出 2007-11-18 19:50:34 補充： 顯示有D問題顯示唔到個加號其實你可以從e-mail問我因為我覺得咁樣會方便D 2007-11-18 20:13:48 補充： 3a)f(x)=3x-1和g(x)=x-2∵2f(a)-3g(a)=5f(a)=2(3a-1) =6a-2g(a)=3(a-2) =3a-6∴6a-2-(3a-6)=53a=1a=1/33b)f(x)=3x-1和g(x)=x-2∵f(2b)-g(3b)=5f(b)=3(2b)-1 =6b-1g(b)=3b-2 ∴6b-1-(3b-2)=53b=4b=4/3 2007-11-18 22:46:15 補充： 4a)f(x)=(x 3)(x 2)-k^2和f(k)=2k∵(k 3)(k 2)-k^2=2k ←代x係kk^2 2k 3k 6-k^2-2k=03k= -6k= -24b) Sorry....我唔識做5b) & 5c)我都唔係好識，我驚教錯你 2007-11-18 22:46:27 補充： 5a)f(x)=x^2-kxf(x 2)=(x 2)^2-k(x 2) =x^2 4x 4-kx-2k =x^2 (4-k)x 4-2kf(x-2))=(x-)^2-k(x-2) =x^2-4x 4-kx 2k =x^2-(4 k)x 4 2k
1.已知f(x)=2(x+m)和g(x)=x+m。若f(8)=6g(4)，求m的值。答案是-2 2(8+m)= 6(4+m) 16+2m=24+6m 4m=-8 m=-2 2.已知f(x)=(x+2)(x-2)+ax+b。若f(2)=2和f(-2)=4,求a和b的值。答案是a=-(1/2),b=3 (2+2)(2-2)+2a+b=2-------1 2a+b=2 (-2+2)(-2-2)-2a+b=-4-----2 2a+b=2 b=2-2a------3 把3代入2... -2a+(2-2a)=4 -2a+2-2a=4 -4a=2 a=-1/2 把a=-1/2代入1.. 2a+b=2 2(-1/2)+b=2 -1+b=2 b=3 3.設f(x)=3x-1和g(x)=x-2。 (a)若2f(a)-3g(a)=5,求a的值。答案是1/3 (b)若f(2b)-g(3b)=5,求b的值。答案是4/3 a. 2(3a-1)-3(a-2)=5 6a-2-3a+6=5 3a=-1 a=-1/3 b. 3(2b)-1-3b-2=5 6b-1-3b-2=5 3b=2 b=2/3 4.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。 (a)求k的值。答案是-2 (b)由此，若2x-f(x)=0,求x的值。答案是-1,-2 a. (k+3)(k+2)-k^2=2k k^2+2k+3k+6-k^2=2k 5k+6=2k 3k=-6 k=-2 b. 2x- (x+3)(x+2) +(-2)^2 = 0 2x -x^2-5x-6+4 = 0 x^2 + 3x + 2 = 0 (x+1)(x+2) = 0 x =-1 or x=-2 5.已知f(x)=x^2-kx。 (a)求f(x+2)和f(x-2)的值。答案是f(x+2)=x^2+(4-k)x+4-2k,f(x-2)=x^2-(4+k)x+4+2k (b)若f(x+2)-f(x-2)=kx-32,求k的值。答案是8 (c)由此，若f(x+2)=f(x-2)+40,求x的值。答案是9 a. f(x+2) = x^2 +(4-k)x + 4 -2k f(x-2) = x^2 -(4+k)x +4 + 2k b. f(x+2) - f(x-2) = 8x - 4k = kx - 32 (x+4)(8-k) = 0 k = 8 c. f(x+2) = f(x-2) + 40 8x - 4k = 40 8x - 4(8) = 40 x = 9|||||1. f(8) = 6g(4) 2(8+m) = 6(4+m) 16+2m = 24+6m -4m= 8 m=-2 2. f(2) = 2a + b = 2 f(-2) = -2a + b = 4 f(2)+f(-2) = 2b = 6 b = 3 2a + 3 = 2 a = -1/2 3. (a) 2f(a) - 3g(a) = 2(3a-1)-3(a-2) 3a + 4 = 5 a= 1/3 (b) f(2b)-g(3b) = 3(2b)-1 -3b +2 = 5 3b +1 = 5 b = 4/3 4. (a)f(k) = (k+3)(k+2)-k^2 = 2k k^2 + 5k +6 - k^2 -2k =0 3k + 6 =0 k = -2 (b) 2x- (x+3)(x+2) +(-2)^2 = 0 2x -x^2-5x-6+4 = 0 x^2 + 3x + 2 = 0 (x+1)(x+2) = 0 x =-1 or x=-2 2007-11-19 22:54:55 補充： 5.(a) f(x+2) = x^2 +(4-k)x + 4 -2k f(x-2) = x^2 -(4+k)x +4 + 2k (b) f(x+2) - f(x-2) = 8x - 4k = kx - 32 (x+4)(8-k) = 0 k = 8 (c) f(x+2) = f(x-2) + 40 8x - 4k = 40 8x - 4(8) = 40 x = 9|||||1.) fx(x+2)+50=fx(x-2)+40-70 =求x的值 2.) 求fx^2(x+2)和fx^2(x-2)的值 等等...
幾題復雜少少二次方程(20點)
發問:
1.已知f(x)=2(x+m)和g(x)=x+m。若f(8)=6g(4)，求m的值。答案是-22.已知f(x)=(x+2)(x-2)+ax+b。若f(2)=2和f(-2)=4,求a和b的值。答案是a=-(1/2),b=33.設f(x)=3x-1和g(x)=x-2。(a)若2f(a)-3g(a)=5,求a的值。答案是1/3(b)若f(2b)-g(3b)=5,求b的值。答案是4/34.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。(a)求k的值。答案是-2(b)由此，若2x-f(x)=0,求x的值。答案是-1,-25.已知f(x)=x^2-kx。(a)求f(x+2)和f(x-... 顯示更多 1.已知f(x)=2(x+m)和g(x)=x+m。若f(8)=6g(4)，求m的值。答案是-2 2.已知f(x)=(x+2)(x-2)+ax+b。若f(2)=2和f(-2)=4,求a和b的值。答案是a=-(1/2),b=3 3.設f(x)=3x-1和g(x)=x-2。 (a)若2f(a)-3g(a)=5,求a的值。答案是1/3 (b)若f(2b)-g(3b)=5,求b的值。答案是4/3 4.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。 (a)求k的值。答案是-2 (b)由此，若2x-f(x)=0,求x的值。答案是-1,-2 5.已知f(x)=x^2-kx。 (a)求f(x+2)和f(x-2)的值。答案是f(x+2)=x^2+(4-k)x+4-2k,f(x-2)=x^2-(4+k)x+4+2k (b)若f(x+2)-f(x-2)=kx-32,求k的值。答案是8 (c)由此，若f(x+2)=f(x-2)+40,求x的值。答案是9 更新: to N.Y Cafe: 第2題中我不太明白你點樣從 「(2) - (1) -2a+b=4 直式加減」 做到 「(-) 2a+b=2 -4a =2 a= -1/2 b=3」 唔該,你可唔可以從(2) - (1)個到做詳細清楚少少呀,多謝指教!!! 更新 2: 4.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。 (b)由此，若2x-f(x)=0,求x的值。答案是-1,-2 我想問個第3頂不是 - 嗎?為何會有個+出現?因為個問題都是-k^2,自然是-(-2)^2.... 2x- (x+3)(x+2) +(-2)^2 = 0
最佳解答:
1) f(x)=2(x+m)和g(x)=x+m f(8)=6g(4) 左方：f(8)=2(8+m) ←代x係8 =16+2m 右方：g(4)=6(4+m) ←代x係4，同埋係6倍！ =24+6m ∵16+2m=24+6m 16-24=6m-2m -8=4m m= -2 2) f(x)=(x+2)(x-2)+ax+b f(2)=2和f(-2)=4 ∵f(2)=2 (2+2)(2-2)+a(2)+b=2 2a+b=2 ←(1) ∵f(-2)=4 (-2+2)(-2-2)+a(-2)+b=4 -2a+b=4 ←(2) (2) - (1) -2a+b=4 直式加減 -) 2a+b=2 -4a =2 a= -1/2 b=3 如果唔介意，我聽日再答埋你其他問題！ 2007-11-18 19:48:52 補充： 好似聯立方程咁呢！只要將D位對返正你應該可以計到嫁(2) - (1) -2a b=4 直式加減-) 2a b=2 -4a =2a= -1/2 ←代返個a入(2a b=2)或代入( -2a b=4)b=3 ←得出 2007-11-18 19:50:34 補充： 顯示有D問題顯示唔到個加號其實你可以從e-mail問我因為我覺得咁樣會方便D 2007-11-18 20:13:48 補充： 3a)f(x)=3x-1和g(x)=x-2∵2f(a)-3g(a)=5f(a)=2(3a-1) =6a-2g(a)=3(a-2) =3a-6∴6a-2-(3a-6)=53a=1a=1/33b)f(x)=3x-1和g(x)=x-2∵f(2b)-g(3b)=5f(b)=3(2b)-1 =6b-1g(b)=3b-2 ∴6b-1-(3b-2)=53b=4b=4/3 2007-11-18 22:46:15 補充： 4a)f(x)=(x 3)(x 2)-k^2和f(k)=2k∵(k 3)(k 2)-k^2=2k ←代x係kk^2 2k 3k 6-k^2-2k=03k= -6k= -24b) Sorry....我唔識做5b) & 5c)我都唔係好識，我驚教錯你 2007-11-18 22:46:27 補充： 5a)f(x)=x^2-kxf(x 2)=(x 2)^2-k(x 2) =x^2 4x 4-kx-2k =x^2 (4-k)x 4-2kf(x-2))=(x-)^2-k(x-2) =x^2-4x 4-kx 2k =x^2-(4 k)x 4 2k
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其他解答:1.已知f(x)=2(x+m)和g(x)=x+m。若f(8)=6g(4)，求m的值。答案是-2 2(8+m)= 6(4+m) 16+2m=24+6m 4m=-8 m=-2 2.已知f(x)=(x+2)(x-2)+ax+b。若f(2)=2和f(-2)=4,求a和b的值。答案是a=-(1/2),b=3 (2+2)(2-2)+2a+b=2-------1 2a+b=2 (-2+2)(-2-2)-2a+b=-4-----2 2a+b=2 b=2-2a------3 把3代入2... -2a+(2-2a)=4 -2a+2-2a=4 -4a=2 a=-1/2 把a=-1/2代入1.. 2a+b=2 2(-1/2)+b=2 -1+b=2 b=3 3.設f(x)=3x-1和g(x)=x-2。 (a)若2f(a)-3g(a)=5,求a的值。答案是1/3 (b)若f(2b)-g(3b)=5,求b的值。答案是4/3 a. 2(3a-1)-3(a-2)=5 6a-2-3a+6=5 3a=-1 a=-1/3 b. 3(2b)-1-3b-2=5 6b-1-3b-2=5 3b=2 b=2/3 4.已知f(x)=(x+3)(x+2)-k^2和f(k)=2k。 (a)求k的值。答案是-2 (b)由此，若2x-f(x)=0,求x的值。答案是-1,-2 a. (k+3)(k+2)-k^2=2k k^2+2k+3k+6-k^2=2k 5k+6=2k 3k=-6 k=-2 b. 2x- (x+3)(x+2) +(-2)^2 = 0 2x -x^2-5x-6+4 = 0 x^2 + 3x + 2 = 0 (x+1)(x+2) = 0 x =-1 or x=-2 5.已知f(x)=x^2-kx。 (a)求f(x+2)和f(x-2)的值。答案是f(x+2)=x^2+(4-k)x+4-2k,f(x-2)=x^2-(4+k)x+4+2k (b)若f(x+2)-f(x-2)=kx-32,求k的值。答案是8 (c)由此，若f(x+2)=f(x-2)+40,求x的值。答案是9 a. f(x+2) = x^2 +(4-k)x + 4 -2k f(x-2) = x^2 -(4+k)x +4 + 2k b. f(x+2) - f(x-2) = 8x - 4k = kx - 32 (x+4)(8-k) = 0 k = 8 c. f(x+2) = f(x-2) + 40 8x - 4k = 40 8x - 4(8) = 40 x = 9|||||1. f(8) = 6g(4) 2(8+m) = 6(4+m) 16+2m = 24+6m -4m= 8 m=-2 2. f(2) = 2a + b = 2 f(-2) = -2a + b = 4 f(2)+f(-2) = 2b = 6 b = 3 2a + 3 = 2 a = -1/2 3. (a) 2f(a) - 3g(a) = 2(3a-1)-3(a-2) 3a + 4 = 5 a= 1/3 (b) f(2b)-g(3b) = 3(2b)-1 -3b +2 = 5 3b +1 = 5 b = 4/3 4. (a)f(k) = (k+3)(k+2)-k^2 = 2k k^2 + 5k +6 - k^2 -2k =0 3k + 6 =0 k = -2 (b) 2x- (x+3)(x+2) +(-2)^2 = 0 2x -x^2-5x-6+4 = 0 x^2 + 3x + 2 = 0 (x+1)(x+2) = 0 x =-1 or x=-2 2007-11-19 22:54:55 補充： 5.(a) f(x+2) = x^2 +(4-k)x + 4 -2k f(x-2) = x^2 -(4+k)x +4 + 2k (b) f(x+2) - f(x-2) = 8x - 4k = kx - 32 (x+4)(8-k) = 0 k = 8 (c) f(x+2) = f(x-2) + 40 8x - 4k = 40 8x - 4(8) = 40 x = 9|||||1.) fx(x+2)+50=fx(x-2)+40-70 =求x的值 2.) 求fx^2(x+2)和fx^2(x-2)的值 等等...
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