Question

[Term Tearm]

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

a) Rút gọn P

b) Chứng minh: Nếu 0<x<1 thì P>0

c) Tìm giá trị lớn nhất của P

Answer 1

[Bùi Thế Hào]

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

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

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

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

=> \(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b/ Nếu 0<x<1 => \(\sqrt{x}-1< 0\); và \(\sqrt{x}>0\)

=> \(P=-\sqrt{x}\left(\sqrt{x}-1\right)>0\)

c/ \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-x+2.\frac{1}{2}\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)

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

=> \(P_{max}=\frac{1}{4}\)

Đạt được khi x=1/4