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標題:

F.5分式方程唔識計(15點!)

發問:

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
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
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