Fatorial #6

Algebra Level 2

$f(x)$ and $g(x)$ are algebraic expressions such that

$\frac{(x!)^2 (x + 2)! (x - 2)!}{(x + 1)! (x - 1)! (x!)^2} = \frac{ f(x) } { g( x) }.$

What is $f(x) - g(x) ?$

Note: $\frac{ f(x) } { g(x) }$ should be simplified to an irreducible positive fraction.

The answer is 3.

1 solution

Fuad Muhammad
Oct 26, 2014

$\frac { { \left( x! \right) }^{ 2 }\left( x+2 \right) !\left( x-2 \right) ! }{ \left( x+1 \right) !\left( x-1 \right) !\left( x! \right) ^{ 2 } } =\frac { a }{ b } ,\quad a-b=?$

From the question, we know that $x!=x\left( x-1 \right) !$

therefore,

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

since $\frac { \left( x+2 \right) }{ \left( x-1 \right) } =\frac { a }{ b }$ is irreducible positive fraction.

so, $a=x+2$ and $b=x-1$

What is the value of $a-b ?$

$a-b=\left( x+2 \right) -\left( x-1 \right) =\boxed { 3 }$

The solution provided is incorrect. It shows poor understanding of Math. If a/b = c/d it doesn't follow a =c and b =d

Roopesh Ranjan - 6 years, 7 months ago

The above solution us correct, because, in the quetion, it was clerly mentioned tha a/b is no more reducable.

Saurav Suman - 6 years, 7 months ago

how can you consider a=x+2 and b=x-1. why not a=2x+4 and b=2x-2 or something like that

Joydeep Ghatak - 6 years, 7 months ago

The problem says a/b is an irreductible positive fraction

Fabio Gama - 6 years, 7 months ago

Fatorial #6

The answer is 3.

1 solution

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