Cho biểu thức P = x/ căn bậc hai x-1 + 2/ căn bậc hai x-2 + 2x - x căn bậc hai x-2/ x -3 căn bậc hai x+2
a) Ta có: P = \(\frac{x}{{\sqrt x - 1}}\) + \(\frac{2}{{\sqrt x - 2}}\) + \(\frac{{2x - x\sqrt x - 2}}{{x - 3\sqrt x + 2}}\)x ³ 0, x\( \ne \)1, x\( \ne \)4
P = \(\frac{x}{{\sqrt x - 1}}\) + \(\frac{2}{{\sqrt x - 2}}\) + \(\frac{{2x - x\sqrt x - 2}}{{x - 3\sqrt x + 2}}\)
P =\(\frac{{x\left( {\sqrt x - 2} \right) + 2\left( {\sqrt x - 1} \right) + 2x - x\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\)
P =\(\frac{{x\sqrt x + 2x - 2\sqrt x - 2 + 2x - x\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\)
P =\(\frac{{2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\) + \(\frac{2}{{\sqrt x - 1}}\)
b)\(\left| P \right|\) – P = 0 Û\(\left| P \right|\) = P Û P > 0
Û\(\frac{2}{{\sqrt x - 1}}\)> 0 Û\(\sqrt x \)> 1 Û x > 1
Kết hợp với ĐKXĐ: x\( \ne \)1, x\( \ne \)4