Question

[Yến Nhi]

Cho M=\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\):(\(\frac{1}{\sqrt{x}+1}-\frac{\sqrt{x}}{1-\sqrt{x}}\)\(+\frac{2}{x-1}\))

a)Rút gọn M

b) so sánh M với 1

c)Tìm gtnn, gtln của M

Answer 1

[Nguyễn Minh Đăng]

a) đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có:

\(M=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\left(\frac{1}{\sqrt{x}+1}-\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{2}{x-1}\right)\)

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

\(M=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}\cdot\frac{x-1}{\sqrt{x}-1+x+\sqrt{x}+2}\)

\(M=\frac{4\sqrt{x}}{x+2\sqrt{x}+1}=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\)

b) Áp dụng BĐT Cauchy ta có: \(\left(\sqrt{x}+1\right)^2\ge4\sqrt{x}\)

\(\Rightarrow M=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\le1\)

Dấu "=" xảy ra khi x = 1 => mâu thuẫn đk

=> \(M< 1\)

c) Vì \(\hept{\begin{cases}4\sqrt{x}\ge0\\\left(\sqrt{x}+1\right)^2>0\end{cases}}\Rightarrow\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\ge0\)

Từ b =>  \(1>M\ge0\)