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## GMAT - Problem Solving - Algebra

LEAP Administrator 10 months ago

### If x ? 0 and x?1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

${ \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }\quad$

If x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

$A.\quad { \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }\\ B.\quad { \left( \frac { x-1 }{ x+1 } \right) }^{ 2 }\\ C.\quad \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \\ D.\quad \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \\ E.-{ \left( \frac { x-1 }{ x+1 } \right) }^{ 2 }\\ \\ \\$

${ \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }\\ replacing\quad x\quad =\quad 1/x\\ =\quad { \left( \frac { \frac { 1 }{ x } +1 }{ \frac { 1 }{ x } -1 } \right) }^{ 2 }\\ ={ \left( \frac { \frac { 1+x }{ x } }{ \frac { 1-x }{ x } } \right) }^{ 2 }\\ since\quad x\quad is\quad not\quad equal\quad to\quad zero\\ ={ \left( \frac { x+1 }{ 1-x } \right) }^{ 2 }\\ since\quad { (1-x })^{ 2 }={ (1+x) }^{ 2 }\\ ={ \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }$