Đáp án: a) P= ($\frac{x-5√x}{x-25}$ - 1) : ($\frac{25-x}{x+2√x -15}$ - $\frac{√x+3}{√x+5}$ + $\frac{√x-5}{√x-3}$
= ($\frac{√x(√x-5)}{(√x-5)(√x+5)}$ - 1) : ($\frac{25-x}{x+5√x-3√x-15}$ - $\frac{√x+3}{√x+5}$ + $\frac{√x-5}{√x-3}$)
= ($\frac{√x}{√x+5}$ - $\frac{√x+5}{√x+5}$) : ( $\frac{25-x}{√x(√x+5)-3(√x+5)}$ - $\frac{√x+3}{√x+5}$ + $\frac{√x-5}{√x-3}$)
= $\frac{√x-(√x+5)}{√x+5}$ : ( $\frac{25-x}{(√x+5)(√x-3)}$ - $\frac{(√x+3)(√x-3)}{√x+5)(√x-3)}$ + $\frac{(√x-5)(√x+5)}{(√x-3)(√x+5)}$)
= $\frac{-5)}{√x+5}$ : $\frac{25-x-(√x²-3²)+(√x²-5²)}{(√x+5)(√x-3)}$
= $\frac{-5}{√x+5}$ . $\frac{(√x+5)(√x-3)}{25-x-x+9+x-25}$
= $\frac{-5(√x-3)}{9-x}$
= $\frac{5(√x-3)}{x-9}$
= $\frac{-5(√x-3)}{(√x-3)(√x+3)}$
= $\frac{5}{√x+3}$
b) Để P < 1 ⇔ $\frac{5}{√x+3}$ < 1
⇔ $\frac{5}{√x+3}$ < $\frac{√x+3}{√x+3}$
⇒ 5 < √x+3
⇔ √x < 2
⇔ x < 4
