Question

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

a) rút gọn P

b) chứng minh rằng nếu 0<x<1 thì P<0

c) tìm giá trị lớn nhất của P

Answer 1

Lời giải:

ĐKXĐ: \(x\neq 1; x\geq 0\)
\(P=\left(\frac{(\sqrt{x}-2)(\sqrt{x}+1)}{(x-1).(\sqrt{x}+1)}-\frac{(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}\right).\frac{(1-x)^2}{2}\)

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

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

b)

Với $0< x< 1$ thì $\sqrt{x}>0; \sqrt{x}< 1\Rightarrow 1-\sqrt{x}> 0$

$\Rightarrow P=\sqrt{x}(1-\sqrt{x})>0$

c)

\(P=\sqrt{x}-x=\frac{1}{4}-(x-\sqrt{x}+\frac{1}{4})=\frac{1}{4}-(\sqrt{x}-\frac{1}{2})^2\leq \frac{1}{4}\)

Vậy $P_{\max}=\frac{1}{4}$. Dấu "=" xảy ra khi $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$

Answer 2

Lời giải:

ĐKXĐ: \(x\neq 1; x\geq 0\)
\(P=\left(\frac{(\sqrt{x}-2)(\sqrt{x}+1)}{(x-1).(\sqrt{x}+1)}-\frac{(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}\right).\frac{(1-x)^2}{2}\)

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

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

b)

Với $0< x< 1$ thì $\sqrt{x}>0; \sqrt{x}< 1\Rightarrow 1-\sqrt{x}> 0$

$\Rightarrow P=\sqrt{x}(1-\sqrt{x})>0$

c)

\(P=\sqrt{x}-x=\frac{1}{4}-(x-\sqrt{x}+\frac{1}{4})=\frac{1}{4}-(\sqrt{x}-\frac{1}{2})^2\leq \frac{1}{4}\)

Vậy $P_{\max}=\frac{1}{4}$. Dấu "=" xảy ra khi $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$