Question

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

a) Rút gọn A

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

Answer 1

ĐK:\(x>0,x\ne1\)

a) \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right].\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right].\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)