Question #53487

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Answer 1 · 2017-11-21T13:49:05

#f_((x)):x^3-4x^2-1=0#

#f_((x)):(2x)/(x-1)-(3x)/(x+1)=1/x#

#x!=0,+-1#

#=>#
#(2x*(x(x+1)(x-1)))/(x-1)-(3x*(x(x+1)(x-1)))/(x+1)=(1*(x(x+1)(x-1)))/x#

#=> 2x^2(x+1)-3x^2(x-1)=x^2-1#

#=> 2x^3+2x^2-3x^3+3x^2=x^2-1#

#=> -x^3+5x^2=x^2-1#

#=> -x^3+4x^2+1=0#

#=> x^3-4x^2-1=0#

#=> x^2(x-4)=1#

#(=>#For # x in RR => lim_(x rarr~ 4.0606)(x^3-4x^2-1)~~0)#