1÷p+1÷q+1÷x=1÷(p +q+x)
plz solve it
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Hi ,
1/p + 1/q + 1/x = 1/( p + q + x )
( qx + px + pq ) /pqx = 1/( p + q + x )
( qx + px + pq )(p + q + x ) = pqx
qx(p+q+x)+px(p+q+x)+pq(p+q+x) =pqx
pqx+q²x+qx²+p² x+pqx+px²+pq+pq²+pqx= pqx
(p+q)x² + (pq+q²+ p²+pq)x+pq+pq² = 0
( p+q)x² +(p²+2pq+ q²) x + pq (1 + q )=0
(p+q )x² + ( p + q )² x + pq( 1 + q ) = 0
(p+q)x² +p(p+q )x+ q( 1+q)x + pq(1+q)=0
(p+q)x[x+p] + q(1 + q)[x+p] = 0
(x + p ) [ x(p+q)+q(1+q) ] = 0
Therefore ,..
x + p = 0 or x(p+q) + q(1+q)= 0
1/p + 1/q + 1/x = 1/( p + q + x )
( qx + px + pq ) /pqx = 1/( p + q + x )
( qx + px + pq )(p + q + x ) = pqx
qx(p+q+x)+px(p+q+x)+pq(p+q+x) =pqx
pqx+q²x+qx²+p² x+pqx+px²+pq+pq²+pqx= pqx
(p+q)x² + (pq+q²+ p²+pq)x+pq+pq² = 0
( p+q)x² +(p²+2pq+ q²) x + pq (1 + q )=0
(p+q )x² + ( p + q )² x + pq( 1 + q ) = 0
(p+q)x² +p(p+q )x+ q( 1+q)x + pq(1+q)=0
(p+q)x[x+p] + q(1 + q)[x+p] = 0
(x + p ) [ x(p+q)+q(1+q) ] = 0
Therefore ,..
x + p = 0 or x(p+q) + q(1+q)= 0
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