已知方程3x²+5x-1=0的两根为x1,x2,不解方程,求下列各式的值。
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3x²+5x-1=0的两根为x1,x2
∴x1+x2=-5/3
x1x2=-1/3
∴①X1X2 =-1/3
②X1*X2 =-1/3
③X1分之1+X2分之1
=(x1+x2)/x1x2
=(-5/3)/(-1/3)
=5
④X1³+X2³
=(x1+x2)(x1²-x1x2+x2²)
=(x1+x2)[(x1+x2)²-3x1x2]
=-5/3(25/9+1)
=-170/27
⑤|X1-X2|
=√(x1-x2)²
= √[(x1+x2)²-4x1x2]=√(25/9+4/3)=√41/3
∴x1+x2=-5/3
x1x2=-1/3
∴①X1X2 =-1/3
②X1*X2 =-1/3
③X1分之1+X2分之1
=(x1+x2)/x1x2
=(-5/3)/(-1/3)
=5
④X1³+X2³
=(x1+x2)(x1²-x1x2+x2²)
=(x1+x2)[(x1+x2)²-3x1x2]
=-5/3(25/9+1)
=-170/27
⑤|X1-X2|
=√(x1-x2)²
= √[(x1+x2)²-4x1x2]=√(25/9+4/3)=√41/3
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