Question

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

a) rút gọn A

b) tìm giá trị lớn nhất của biểu thức P=A-9\(\sqrt{x}\)

Answer 1

Bài làm :

ĐKXD : \(x>0\) ; \(x\ne1\)

Rút gọn :

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

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

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

\(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

Vậy \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)