Ta có:
\(\displaystyle A= {1 \over x} + {1 \over {x + 5}} + {{x - 5} \over {x\left( {x + 5} \right)}}\)\(\displaystyle = {{x + 5} \over {x\left( {x + 5} \right)}} + {x \over {x\left( {x + 5} \right)}} + {{x - 5} \over {x\left( {x + 5} \right)}}\)
\(\displaystyle = {{x + 5 + x + x - 5} \over {x\left( {x + 5} \right)}} \)\(\displaystyle = {{3x} \over {x\left( {x + 5} \right)}} = {3 \over {x + 5}}=B\)
Vậy \(\displaystyle A = B.\)
