Question

[Nguyễn Thị Thanh Mai]

Cho y = \(\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}\) + 1 - \(\frac{2x+\sqrt{x}}{\sqrt{x}}\)

a. Rút gọn y. Tìm x để y = 2

b. Cho x > 1. Chứng minh y - |y| = 0

c. Tìm GTNN của y

Answer 1

[Trần Đăng Nhất]

a/ ĐKXĐ: \(x>0\)

\(y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)

\(=\frac{\sqrt{x}\left[\left(\sqrt{x}\right)^3+1\right]}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

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

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

Ta có: \(y=2\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=1\end{matrix}\right.\)

Vậy ...............

b/ Ta có: \(x>1\Rightarrow\sqrt{x}>1\Rightarrow\sqrt{x}-1>0\) và \(\sqrt{x}>0\)

nên \(y=\sqrt{x}\left(\sqrt{x}-1\right)>0\). Khi đó \(\left|y\right|=y\)

\(\Rightarrow y-\left|y\right|=y-y=0\) (ĐPCM)

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

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

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)

Vậy \(Min_y=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)