已知:x3−3x2+3x+7=0x^3 - 3x^2 + 3x + 7 = 0x3−3x2+3x+7=0
試求:
∣x+1∣|x+1|∣x+1∣ 與 lnx\ln xlnx
cosx\cos xcosx
方程式可分解為:
(x+1)(x2−4x+7)=0(x + 1) (x^2 - 4 x + 7) = 0(x+1)(x2−4x+7)=0
解得:
x=−1,2±3ix=-1, 2\pm\sqrt{3}ix=−1,2±3i
因此:
xxx
∥x+1∥\|x+1\|∥x+1∥
lnx\ln xlnx
-1
0
iπi\piiπ
cos(1)\cos(1)cos(1)
2+3i2+\sqrt{3}i2+3i
232\sqrt{3}23
ln7+tan−1(32)\ln\sqrt{7}+\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)ln7+tan−1(23)
cos(2)cosh(3)−isin(2)sinh(3)\cos(2)\cosh(\sqrt{3})-i \sin(2)\sinh(\sqrt{3})cos(2)cosh(3)−isin(2)sinh(3)
2−3i2-\sqrt{3}i2−3i
ln7−tan−1(32)\ln\sqrt{7}-\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)ln7−tan−1(23)
cos(2)cosh(3)+isin(2)sinh(3)\cos(2)\cosh(\sqrt{3})+i \sin(2)\sinh(\sqrt{3})cos(2)cosh(3)+isin(2)sinh(3)
參考:
Wolfram Alpha:解方程式 x3−3x2+3x+7=0x^3 - 3x^2 + 3x + 7 = 0x3−3x2+3x+7=0
Wolfram Alpha:計算 ln(−1)\ln(-1)ln(−1)
Wolfram Alpha:計算 ln(2+3i)\ln (2+\sqrt{3}i)ln(2+3i)
Wolfram Alpha:計算 cos(2+3i)\cos(2+\sqrt{3}i)cos(2+3i)
Last updated 5 years ago
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