MATHS F.4......THX!!!

102
x^9=3x^5+40x
x^9－3x^5－40x=0
x(x^8－3x^4－40)=0
x(x^4+5)(x^4－8)=0
x^4+5>0
x(x^4－8)=0
x(x^2－8^(1/2))(x^2+8^(1/2))=0
x(x^2－8^(1/2))=0
x(x－8^(1/4)(x+8^(1/4)=0
x=0or x=8^(1/4) or x=－8^(1/4)

103
1/(x+1)^2－1/(x+1)=－1/4
4－4(x+1)=－(x+1)^2，x<>－1
x^2+2x+1+4-4x-4=0，x<>－1
x^2－2x+1=0，x<>－1
x=1double

104
2√(x+1)-6/√(2x－1)=0
x+1=9/(2x－1)
2x^2+x－1=9
2x^2+x－10=0
(2x+5)(x－2)=0
x=-5/2(不合)orx=2

105(b)
(x^2－2x)^2－7x^2+14x－8=0
Sety=x^2－2x
y^2－7y－8=0
(y－8)(y+1)=0
y=8or y=－1
x^2－2x=8or x^2－2x=－1
x^2－2x-8=0or x^2－2x+1=0
(x－4)(x+2)=0or (x－1)^2=0
x=4or x=－2 or x=1(double)

106(a)
log8*log2=3log2*log2=3(log2)^2
(b)log2_x-logx_8=2
logx/log2-log8/logx=2
logx/log2-3log2/logx=2
log2*logx*[logx/log2-3log2/logx]=2log2*logx
(logx)^2-3(log2)^2=2log2*logx
(logx)^2-2log2*log-3(log2)^2=0
(logx+log2)(logx-3log2)=0
logx=-log2orlogx=3log2
x=1/2or x=8

107 (a)
4^3+k*(2^4)+48=0
16k=－112
k=－7
(b)
f(x)=4^x-7*(2^(x+1))+48=0
Set y=2^x
y^2－14y+48=0
(y－6)(y－8)=0
y=6 or y=8
2^x=6 or 2^x=8
x=log2_6or x=3

109 (a)
√(x^2+5x+1)
u^2=x^2+5x+1
3u^2+2u－3
=3x^2+15x+3+2√(x^2+5x+1)－3
=3x^2+15x+2√(x^2+5x+1)
(b)
3u^2+2u－3=2
3u^2+2u－5=0
(3u+5)(u－1)=0
u=－5/3(不合) or u=1
√(x^2+5x+1)=1
x^2+5x+1=1
x(x+5)=0
x=0or x=－5

109 (a)
(x+√(x+2))^2－2
=x^2+2x√(x+2)+(x+2)－2
=x^2+x+2x√(x+2)

(b) x^2+x+2x√(x+2)=14
(x+√(x+2))^2－2=14
(x+√(x+2))^2=16
x+√(x+2)=4 or x+√(x+2)=－4
x+√(x+2)=4
√(x+2)=4－x
x+2=x^2－8x+16
x^2－9x+14=0
(x－2)(x－7)=0
x=2or x=7(驗算不符)
x+√(x+2)=－4
√(x+2)=－4－x
x+2=x^2+8x+16
x^2+7x+14=0
D=7^2－4*1*14<0(不合)



2010-08-09 11:19:04 補充：
110
(1/x－1)(1/x+2)=4
(1－x)(1+2x)=4x^2，x<>0
-2x^2+x+1=4x^2，x<>0
6x^2－x－1=0，x<>0
(3x+1)(2x－1)=0，x<>0
x=－1/3 or x=1/2

112.
k^2=k+3
k^4－2k^3+k^2=k^2(k^2－2k+1)=(k+3)(k+3－2k+1)=(k+3)(4－k)
=－k^2+k+12=－k－3+k+12=9

102.)
x^9 = 3x^5 + 40x
x^8 = 3x^4 + 40
x^8 - 3x^4 - 40 = 0
(x^4 - 8)(x^4 + 5) = 0
x^4 = 8 or -5(rej.)
x^4 - 8 = 0
(x^2 - 2√2)(x^2 +2√2) = 0
x^2 = 2√2 or -2√2(rej.)
x = 1.68 or -1.68 (corr to 3 sig fig.)
103.)
1/(x+1)^2 - 1/(x+1) = -1/4
4 - 4(x+1) = -(x+1)^2
(x+1)^2 - 4(x+1) + 4 = 0
[(x+1) - 2]^2 = 0
x+1-2=0
x-1 = 0
x = 1
104.)
2√(x+1) - 6/√(2x-1) = 0
√(x+1) - 3/√(2x-1) = 0
√(x+1)(2x-1) - 3 = 0
(x+1)(2x-1) = 9
2x^2 +x - 10 = 0
(2x+5)(x-2) = 0
x = -5/2 or 2
Since -5/2 + 1 < 0, hence x = 2 only. < x= -5/2 is rejected.
105b.)
Let (x^2 - 2x) be y,
(x^2 - 2x)^2 - 7x^2 + 14x = 8
(x^2 - 2x)^2 - 7(x^2 - 2x) - 8 = 0
y^2 - 7y - 8 = 0
(y-8)(y+1) = 0
y = 8 or -1
x^2 - 2x = 8 or -1
x^2 - 2x - 8 = 0 or x^2 - 2x + 1 = 0
(x-4)(x+2) = 0 or (x-1)^2 = 0
x = 4 or -2 or 1
106a.)
log8 * log 2
= log2^3 * log 2
= 3log2 * log 2
= 3(log2)^2
b.)
log_2 x - log_x 8 = 2
logx/log2 - log8/logx = 2
(logx * logx)/(log2 * log 8) = 2
(logx)^2 = 6(log2)^2
(logx)^2 = (log2^3)(log2^3)
(logx)^2 = (log8)^2
(logx-log8)(logx+log8) = 0
logx = log8 or log x = -log8
x = 8 or 1/8
107a.)
f(3)=0, 4^3 + k(2^4)+48 = 0
k(16)=-112
k=-7
b.)
f(x) = 0,
4^x - 7(2^(x+1))+48=0
2^2x - 14(2^x) + 48 = 0
(2^x - 6)(2^x - 8) = 0
2^x = 6 or 8
x = 2.58 or 3 (corr to 3 sig fig)
108.a)
Given √(x^2 + 5x + 1) = u,
3x^2 + 15x + 2√(x^2 + 5x + 1)
= 3x^2 + 15x + 3 + 2√(x^2 + 5x + 1) - 3
= 3(x^2 + 5x + 1)+2√(x^2 + 5x + 1) - 3
= 3u^2 + 2u - 3
b.)3x^2 + 15x + 2√(x^2 + 5x + 1) = 2,
3u^2 + 2u - 3 = 2
3u^2 + 2u - 5 = 0
(3u+5)(u-1) = 0
u = -5/3 or 1
√(x^2 + 5x + 1) = -5/3(rej.) or 1
x^2 + 5x + 1 = 1
x^2 + 5x = 0
x(x+5) = 0
x = 0 or -5
109a.)
L.H.S. = x^2 + x +2x√(x+2)
= x^2 + 2x√(x+2) + (x+2) - 2
= [x+√(x+2)]^2 -2
= R.H.S.
b.) x^2 + x +2x√(x+2) = 14
[x+√(x+2)]^2 -2 = 14
[x+√(x+2)]^2 = 16
x+√(x+2) = 4 or -4
x+√(x+2) -4 = 0 or x+√(x+2) + 4 = 0
(x+2)+√(x+2)-6=0 or (x+2)+√(x+2)+2=0(rej.)
[√(x+2) +3][√(x+2) - 2] = 0
√(x+2) = -3(rej.) or 2
x+2 = 4
x = 2
110.)
(1/x -1)(1/x +2) = 4
(1/x)^2 + 1/x - 6 = 0
[(1/x)+3][(1/x)-2] = 0
1/x = -3 or 1/x = 2
x = -1/3 or 1/2




2010-08-08 17:15:10 補充：
112.)
k is a root of x^2-x-3=0,
k^2 - k - 3=0
[k^2 - (k+3)]^2 =0
k^4 - 2k^2 (k+3) + (k+3)^2 = 0
k^4 - 2k^3 - 6k^2 + k^2 + 6k + 9 = 0
k^4 - 2k^3 - 5k^2 + 6k + 9 = 0
k^4 - 2k^3 - 5k^2 + 6k + 9 +6(k^2 - k - 3) = 0
k^4 - 2k^3 - 5k^2 + 6k + 9 + 6k^2 - 6k - 18 = 0
k^4 - 2k^3 + k^2 = 9

2010-08-08 17:19:36 補充：
106b 不太確定~_~"