已知代数式3y的2次方-2y+6的值为8,那么代数式3/2y-y+1的值为多少?
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解:
3y^2-2y+6=8
3y^2-2y-2=0
y=[2±√(4+4*3*2)]/(2*3)=(1±√7)]/3
y1=(1+√7)/3≠0
y2=(1-√7)/3≠0
设(3/2y-y+1)=L,则把y1,y2的值代入得
L1=(-1+5√7)/12
L2=-(1+5√7)/12
或L=(3/2y-y+1)=(3-2y^2+2y)
=-(3y^2-3y-4.5)/(3y)
=-(0-y-2.5)/(3y)
=(y+2.5)/(3y)
=1/3+2.5/(3y)
L1=1/3+2.5/[3*(1+√7)/3]
=(-1+5√7)/12
L2=1/3+2.5/[3*(1-√7)/3]
=-(1+5√7)/12
3y^2-2y+6=8
3y^2-2y-2=0
y=[2±√(4+4*3*2)]/(2*3)=(1±√7)]/3
y1=(1+√7)/3≠0
y2=(1-√7)/3≠0
设(3/2y-y+1)=L,则把y1,y2的值代入得
L1=(-1+5√7)/12
L2=-(1+5√7)/12
或L=(3/2y-y+1)=(3-2y^2+2y)
=-(3y^2-3y-4.5)/(3y)
=-(0-y-2.5)/(3y)
=(y+2.5)/(3y)
=1/3+2.5/(3y)
L1=1/3+2.5/[3*(1+√7)/3]
=(-1+5√7)/12
L2=1/3+2.5/[3*(1-√7)/3]
=-(1+5√7)/12
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