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F.5分式方程唔識計(15點!)
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Question 1 1/x + 1 +2/x = 3 1/x + 2/x = 2 (1+2)/x = 2 x = 3/2 Question 2 (5^x)^2 - 7(5^x) + 6 = 0 ((5^x) - 6)((5^x)-1) = 0 Therefore 5^x = 6 or 5^x = 1 If 5^x = 6, then log 5^x = log 6 x (log 5) = log 6 x = log 6/ log 5 = 1.113 If 5^x = 1, then log 5^x = log 1 x (log 5) = log 1 x = log 1/ log 5 = 0 Question 3 Log (x+2) + log x = 1 Log [(x+2)(x)] = log 10 x^2+2x = 10 x^2 +2x -10 = 0 <== Can't be solved by factorization method, try using completing square method or your programmed calculator. Question 4 4cos^2θ=3cosθ 4cos^2θ - 3cosθ = 0 cosθ (4 cosθ - 3) = 0 cos θ = 0 or cos θ = 3/4 if cos θ = 0, then θ = 90 Deg or 270 Deg; if cos θ = 3/4, then θ = 41.40 Deg or 318.60 Deg. Question 5 6sin^2θ-sinθ-1=0 (3sinθ+1)(2sinθ-1) = 0 sinθ = -1/3 or sinθ = 1/2 if sinθ = -1/3, then θ = 340.53 Deg or 199.47 Deg; if sinθ = 1/2, then θ = 30 Deg or 150 Deg. Question 6 2x+y =5 (Given condition 1) y= -2x + 5 x + y^2 = 10 (Given condition 2) Substiting y = -2x + 5 into equation from condition 2 x + (-2x + 5)^2 = 10 x + (4x^2- 20x + 25) =10 4x^2 -19x +15 = 0 (4x -15)(x-1) = 0 x = 15/4 (rejected) or x = 1 y = (-2)(1)+ 5 = 3 2010-09-18 18:33:06 補充： Question 3 Log (x+2) + log x = 1 Log [(x+2)(x)] = log 10 x^2+2x = 10 x^2 +2x -10 = 0 <== Can't be solved by factorization method, try using completing square method or your programmed calculator. x=2.32 or x = -4.32 2010-09-18 18:36:40 補充： x = -4.32 (Rejected) 2010-09-18 18:41:07 補充： For Question 1, if your question is 1/x + 1 +2/x = 3, please take my solution; if your question is 1/(x + 1) +2/x = 3, please take yeung's solution.
其他解答:
1/x+1+2/x=31/(x+1) +2/x=3X+2(x+1)=3x(x+1)X+2x+2=3x^2+3x3x^2-2=0X=√(2/3) (5^x)^2-7(5^x)+6=0[(5^x)-6][(5^x)-1]=05^x-6=0 or 5^x-1=0Xlog5=log6 or xlog5=log1X=1.113 or x=0 log(x+2)+logx=1log[(x+2)x]=1x^2+2x=10x^2+2x-10=0x={-2+√[2^2-4(-10)] }/2 or x={-2-√[2^2-4(-10)] }/2x=(-2+2√11)/2 or x=(-2-2√11)/2x=-1+ √11 or x =-1-√11 (rejected x<0)x=2.317 以下是三角方程,其中0 ≤ θ < 360 !!!!! 4cos^2θ=3cosθcosθ(4cosθ-3)=0cosθ=0 or cosθ=3/4θ=90 or 270 or 41.41 or 318.59 6sin^2θ-sinθ-1=0(3sinθ+1)(2sinθ-1)=0Sinθ=-1/3 or sinθ=1/2Θ=340.53 or 199.47 or 30 or 1504E350C6F99F7EF4C
F.5分式方程唔識計(15點!)
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1/x+1+2/x=3 5^x)^2-7(5^x)+6=0 log(x+2)+logx=1 以下是三角方程,其中0 ≤ θ < 360 !!!!! 4cos^2θ=3cosθ 6sin^2θ-sinθ-1=0 x和y是兩個正整數,x的兩倍加上y是5,而x加上y的平方則是10 i)試以聯立方程表示x和y的關係 ii)利用代數方法求x和y的值最佳解答:
Question 1 1/x + 1 +2/x = 3 1/x + 2/x = 2 (1+2)/x = 2 x = 3/2 Question 2 (5^x)^2 - 7(5^x) + 6 = 0 ((5^x) - 6)((5^x)-1) = 0 Therefore 5^x = 6 or 5^x = 1 If 5^x = 6, then log 5^x = log 6 x (log 5) = log 6 x = log 6/ log 5 = 1.113 If 5^x = 1, then log 5^x = log 1 x (log 5) = log 1 x = log 1/ log 5 = 0 Question 3 Log (x+2) + log x = 1 Log [(x+2)(x)] = log 10 x^2+2x = 10 x^2 +2x -10 = 0 <== Can't be solved by factorization method, try using completing square method or your programmed calculator. Question 4 4cos^2θ=3cosθ 4cos^2θ - 3cosθ = 0 cosθ (4 cosθ - 3) = 0 cos θ = 0 or cos θ = 3/4 if cos θ = 0, then θ = 90 Deg or 270 Deg; if cos θ = 3/4, then θ = 41.40 Deg or 318.60 Deg. Question 5 6sin^2θ-sinθ-1=0 (3sinθ+1)(2sinθ-1) = 0 sinθ = -1/3 or sinθ = 1/2 if sinθ = -1/3, then θ = 340.53 Deg or 199.47 Deg; if sinθ = 1/2, then θ = 30 Deg or 150 Deg. Question 6 2x+y =5 (Given condition 1) y= -2x + 5 x + y^2 = 10 (Given condition 2) Substiting y = -2x + 5 into equation from condition 2 x + (-2x + 5)^2 = 10 x + (4x^2- 20x + 25) =10 4x^2 -19x +15 = 0 (4x -15)(x-1) = 0 x = 15/4 (rejected) or x = 1 y = (-2)(1)+ 5 = 3 2010-09-18 18:33:06 補充： Question 3 Log (x+2) + log x = 1 Log [(x+2)(x)] = log 10 x^2+2x = 10 x^2 +2x -10 = 0 <== Can't be solved by factorization method, try using completing square method or your programmed calculator. x=2.32 or x = -4.32 2010-09-18 18:36:40 補充： x = -4.32 (Rejected) 2010-09-18 18:41:07 補充： For Question 1, if your question is 1/x + 1 +2/x = 3, please take my solution; if your question is 1/(x + 1) +2/x = 3, please take yeung's solution.
其他解答:
1/x+1+2/x=31/(x+1) +2/x=3X+2(x+1)=3x(x+1)X+2x+2=3x^2+3x3x^2-2=0X=√(2/3) (5^x)^2-7(5^x)+6=0[(5^x)-6][(5^x)-1]=05^x-6=0 or 5^x-1=0Xlog5=log6 or xlog5=log1X=1.113 or x=0 log(x+2)+logx=1log[(x+2)x]=1x^2+2x=10x^2+2x-10=0x={-2+√[2^2-4(-10)] }/2 or x={-2-√[2^2-4(-10)] }/2x=(-2+2√11)/2 or x=(-2-2√11)/2x=-1+ √11 or x =-1-√11 (rejected x<0)x=2.317 以下是三角方程,其中0 ≤ θ < 360 !!!!! 4cos^2θ=3cosθcosθ(4cosθ-3)=0cosθ=0 or cosθ=3/4θ=90 or 270 or 41.41 or 318.59 6sin^2θ-sinθ-1=0(3sinθ+1)(2sinθ-1)=0Sinθ=-1/3 or sinθ=1/2Θ=340.53 or 199.47 or 30 or 1504E350C6F99F7EF4C
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