a) P = $\frac{1}{2√x - 2}$ - $\frac{1}{2√x + 2}$ + $\frac{√x}{1 - x}$ Đkxđ: x ≥ 0; x $\neq$ 1
= $\frac{1}{2(√x - 1)}$ - $\frac{1}{2(√x + 1)}$ - $\frac{√x}{x- 1}$
= $\frac{1.(√x + 1) - 1.(√x - 1) - 2.√x}{2(√x + 1)(√x - 1)}$
= $\frac{√x + 1 - √x + 1 - 2√x}{2(√x + 1)(√x - 1)}$
= $\frac{-2√x + 2}{2(√x + 1)(√x - 1)}$
= $\frac{-2(√x - 1)}{2(√x + 1)(√x - 1)}$
= $\frac{-1}{√x + 1}$
b) Có: x = $\frac{4}{9}$ (t/m)
Thay x = $\frac{4}{9}$ vào P ta được:
P = $\frac{-1}{\sqrt[2]{\frac{4}{9}} + 1}$
= $\frac{-1}{\frac{2}{3} + 1}$
= $\frac{-1}{\frac{5}{3}}$
= -$\frac{3}{5}$
Vậy P = -$\frac{3}{5}$ khi x = $\frac{4}{9}$
c) |P| = $\frac{1}{3}$ Đkxđ: x ≥ 0; x $\neq$ 1
⇔ |$\frac{-1}{√x + 1}$| = $\frac{1}{3}$
TH1: $\frac{-1}{√x + 1}$ = $\frac{1}{3}$
⇔ √x + 1 = -3
⇔ √x = -4
Mà √x ≥ 0 với mọi x ∈ đkxđ
⇒ x ∈ ∅
TH2: $\frac{-1}{√x + 1}$ = -$\frac{1}{3}$
⇔ √x + 1 = 3
⇔ √x = 2
⇔ x = 4 (t/m)
Vậy x = 4 thì |P| = $\frac{1}{3}$
