-X∧4+5X∧3-4X∧2-5X-1怎么求?-X∧4+5X∧3-4X∧2-5X-1=0
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-X∧4+5X∧3-4X∧2-5X-1怎么求?-X∧4+5X∧3-4X∧2-5X-1=0
-X∧4+5X∧3-4X∧2-5X-1怎么求?
-X∧4+5X∧3-4X∧2-5X-1=0
-X∧4+5X∧3-4X∧2-5X-1怎么求?-X∧4+5X∧3-4X∧2-5X-1=0
-(X∧4-5X∧3+4X∧2+5X+1)=0
X∧4-5X∧3+4X∧2+5X+1=0
令(X^2+AX-1)(X^2+BX-1)=X∧4-5X∧3+4X∧2+5X+1
左面展开得:
X^4+(A+B)X^3+(AB-2)X^2-(A+B)X+1=X∧4-5X∧3+4X∧2+5X+1
比较系数,得到:
A+B=-5,AB-2=4
解方程组,得:
A1=-1,B1=-4;A2=-4,B2=-1
所以,X∧4-5X∧3+4X∧2+5X+1
=(X^2-X-1)(X^2-4X-1)
=0
当(X^2-X-1) =0时,
X1,X2=(1±√5)/2
当(X^2-4X-1)=0时,
X3,X4=2±√17/2
所以,原方程有四个
X1=(1+√5)/2
X2=(1-√5)/2
X3=2+√17/2
X4=2-17/2
不是方程,无法求出x
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
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