Question

P = \(\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\frac{1-\sqrt{x}}{1-x}\) với x\(\ge\)0 và x # 1
a) Rút gọn P
b) Tính P khi x = \(\sqrt{\frac{5}{2}-\sqrt{6}}\)

Answer 1

\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}+\sqrt{x}\right)\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)

\(=\left(x+2\sqrt{x}+1\right)\frac{1}{\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}=\sqrt{x}+1\)

\(x=\sqrt{\frac{5}{2}-\sqrt{6}}=\sqrt{\frac{5-2\sqrt{6}}{2}}=\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{2}}=\frac{\sqrt{6}-2}{2}\)

\(\Rightarrow P=\sqrt{\frac{\sqrt{6}-2}{2}}+1\)