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Question Number 152740 by mr W last updated on 31/Aug/21

Commented by mr W last updated on 31/Aug/21

[Q152712]

[Q152712]

Answered by mr W last updated on 31/Aug/21

Commented by mr W last updated on 31/Aug/21

AF=CL  EO=((CL)/2)  LM=(2/3)×EO=((CL)/3)  CM=(2/3)×CL   ⇒((CL)/(CM))=(3/2)  GF=((CN)/2)  MN=((AG)/3) ⇒AG=3×MN  AG+GF=AF=CL  3×MN+((CN)/2)=CL   ...(i)  MN+CN=(2/3)×CL   ...(ii)  3×(ii)−(i):  (5/2)×CN=CL  ⇒((CL)/(CN))=(5/2)  ((CL)/(CL))+((CL)/(CM))=1+(3/2)=(5/2)=((CL)/(CN))  ⇒(1/(CL))+(1/(CM))=(1/(CN))

AF=CLEO=CL2LM=23×EO=CL3CM=23×CLCLCM=32GF=CN2MN=AG3AG=3×MNAG+GF=AF=CL3×MN+CN2=CL...(i)MN+CN=23×CL...(ii)3×(ii)(i):52×CN=CLCLCN=52CLCL+CLCM=1+32=52=CLCN1CL+1CM=1CN

Commented by Tawa11 last updated on 31/Aug/21

God bless you sir. I appreciate sir.

Godblessyousir.Iappreciatesir.

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