Question

Cho biểu thức P= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{2}{x-1}+\dfrac{1}{\sqrt{x}+1}\right)\)với x>0, x≠1

a. Rút gọn P

b. Tìm x để P= \(\dfrac{1}{2}\), P= -1

c. Tìm số nguyên x để P nhận giá trị là số nguyên

Answer 1

P=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{2}{x-1}+\dfrac{1}{\sqrt{x}+1}\right)\)

=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{2}{x-1}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

=\(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

=\(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)