Question

[Hoàng Thị Mai Trang]

Cho biểu thức:P=\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\) với x>0;x≠1

a)Rút gọn P

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

c)Tìm x để P có GTLN

Answer 1

[Hồng Phúc]

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

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

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

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

b, \(x=7-4\sqrt{3}=\left(\sqrt{3}-2\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)

Khi đó \(P=-x+\sqrt{x}=-\left(7-4\sqrt{3}\right)+\left(2-\sqrt{3}\right)=-5+3\sqrt{3}\)

c, \(P=-x+\sqrt{x}=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(\Rightarrow MaxP=\frac{1}{4}\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)