RRB Junior Engineer Stage 1 Model Papers 6

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Question : 10 of 99

Marks: +1, -0

If

xy+yz+zx=0

, then

(1/{x^2-yz}+1/{y^2-zx}+1/{z^2-xy})(x,y,z≠0)

is equal to

0
3
1
$x + y + z$

Solution:

xy+yz+zn=0

1/{x^2-yz}+1/{y^2-zx}+1/{z^2-xy}

=1/{x^2+xy+zn}+1/{y^2+xy+yz}+1/{z^2+yz+xz}

=1/{x(x+y+z)}+1/{y(x+y+z)}+1/{z(x+y+z)}

={yz+xz+xy}/{xyz(x+y+z)}=0/{{xyz(x+y+z)}}=0

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