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题：求 $x(x+1)(x+2)(x+3)(x+4)=120$ 的根式解.B@
解：$x(x+1)(x+2)(x+3)(x+4)-120$P
$\qquad=(x-1)(x^4+11x^3+46x^2+96x+120)$OXPDo
$\;\;$设$\;\,x^4+11x^3+46x^2+96x+120=(x^2+\frac{11}{2}x+A)^2-(Bx+C)^2$Q
$\;\;\;=x^4+11x^3+(2A-B^2+\frac{121}{4})x^2+(11A-2BC)x+A^2-C^2$_ze3n&
$\;\;\;=(x^2{\small+}(\frac{11}{2}{\small+B})x{\small+A+C})(x^2{\small+}(\frac{11}{2}{\small-B})x{\small+A-C})$Z
$\;\;$则$\;B=\frac{11A-96}{2C},\;\;C^2=A^2-120,\;\;2A-\big(\frac{11A-96}{2C}\big)^2+\frac{121}{4}-46=0.$%zO\
$\;\;$整理得${\small\;A^3-23A^2+144A-207=0,\;\;A=}{\large\frac{23}{3}}+{\large\frac{2}{3}}{\small\text{Re}}\sqrt[{\Large 3}]{\large\frac{115+i9\sqrt{44907}}{2}}$[9O=f!
$\;\;$是上方程使$\small\,B,\,C\,$皆为实数的唯一解.8Q,8
$\;\;$令$\;\theta=\arctan{\large\frac{9\sqrt{44907}}{115}},\;\;u=\sqrt{97}\cos\large\frac{\theta}{3},\;\;$则$\;A={\large\frac{23}{3}}+{\large\frac{2}{3}}u.$z
$\;\;\small B=\sqrt{2A+(11/2)^2-46}={\large\frac{\sqrt{48u-15}}{6}},\;C={\large\frac{11A-96}{2B}}={\large\frac{22u-35}{\sqrt{48u-15}}}$8<z
$\;\;$因此原方程的根为$\;x_0=1,$N>}+ %
$\qquad x_{1,2}={\large\frac{-(\frac{11}{2}+{\small B})\pm\sqrt{(\frac{11}{2}+{\small B})^2\small-4(A+C)}}{2}}$!B#
$\qquad\qquad={\large\frac{-33-\sqrt{48u-15}\pm i\sqrt{6}\sqrt{8u+5-\frac{225\sqrt{3}}{\sqrt{16u-5}}}}{12}}\underset{\,}{\,}$]?9S]
$\qquad\qquad=-4.41473149115\ldots\pm i1.29458342150\ldots\underset{\,}{\,}$=ZSO
$\qquad x_{3,4}={\large\frac{-(\frac{11}{2}-{\small B})\pm\sqrt{(\frac{11}{2}+{\small B})^2\small-4(A-C)}}{2}}$3,
$\qquad\qquad={\large\frac{-33-\sqrt{48u-15}\pm i\sqrt{6}\sqrt{8u+5+\frac{225\sqrt{3}}{\sqrt{16u-5}}}}{12}}\underset{\,}{\,}$+s
$\qquad\qquad=-1.08526850885\ldots\pm i2.11936680167\ldots$Hmv"VH
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2021/01/02 03:39am 
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