Question

Cho biểu thức:A=\(\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}\right):\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}\)

a)Rút gọn A

b)Tính giá trị của A khi x=7+4\(\sqrt{3}\)

c)Với giá trị nào của x thì A đạt giá trị nhỏ nhất

Answer 1

a)

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

⇔\(\left[\dfrac{\left(1+\sqrt{x}\right)+\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}:\dfrac{\left(1+\sqrt{x}\right)-\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right]+\dfrac{1}{1-\sqrt{x}}\)

⇔\(\left[\dfrac{\left(1+\sqrt{x}\right)+\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}.\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)-\left(1-\sqrt{x}\right)}\right]+\dfrac{1}{1-\sqrt{x}}\)

⇔\(\left[\dfrac{\left(1+\sqrt{x}\right)+\left(1+\sqrt{x}\right)}{\left(1+\sqrt{x}\right)-\left(1+\sqrt{x}\right)}\right]\dfrac{1}{1-\sqrt{x}}\)

⇔\(\left[\dfrac{1+\sqrt{x}+1-\sqrt{x}}{1+\sqrt{x}-1+\sqrt{x}}\right]+\dfrac{1}{1-\sqrt{x}}\)

⇔\(\dfrac{2}{2\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\)

⇔\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\)

⇔\(\dfrac{\sqrt{x}}{\sqrt{x}-x}+\dfrac{\sqrt{x}}{\sqrt{x}-x}\)

⇔\(\dfrac{2\sqrt{x}}{2\sqrt{x}-2x}\)

⇔\(\dfrac{1}{-2x}\)

b) Thay x = \(7+4\sqrt{3}\) vào A ta có:

A = \(\dfrac{1}{-2\left(7+4\sqrt{3}\right)}\)

=\(\dfrac{-7+4\sqrt{3}}{2}\)

Mình chỉ làm dược câu a và câu b thôi, mong bạn thông cảm