Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
\(A=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left[\left(\frac{1}{2\sqrt{x}}\right)^2-2.\frac{1}{2\sqrt{x}}.\frac{\sqrt{x}}{2}+\left(\frac{\sqrt{x}}{2}\right)^2\right]\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right]\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)\)
\(\Leftrightarrow A=\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\left(\frac{1}{4x}-\frac{2x}{4x}+\frac{x^2}{4x}\right)\)
\(\Leftrightarrow A=\frac{-4\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{4\sqrt{x}}{1-x}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{1-x}{\sqrt{x}}\)
b) \(\frac{A}{\sqrt{x}}>1\)
\(\Leftrightarrow\frac{1-x}{\frac{\sqrt{x}}{\sqrt{x}}}>1\)
\(\Leftrightarrow1-x>1\Leftrightarrow x< 0\)
1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)
Đạt được khi x = 9
2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)
Không có GTLN nhé
Lời giải:
ĐK: $x\geq 0; x\neq 1$
$P=\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$
$=\frac{1}{\sqrt{x}-1}=-\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$
$=\frac{x+\sqrt{x}+1-(x+2)-(x-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$
$=\frac{-\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}$
$\Rightarrow Q=\frac{2(x+\sqrt{x}+1)}{-\sqrt{x}}+\sqrt{x}$
$=-\left(\sqrt{x}+\frac{2}{\sqrt{x}}+2\right)$
Dễ thấy $\sqrt{x}+\frac{2}{\sqrt{x}}+2\geq 2\sqrt{2}+2$ theo BĐT Cô-si
$\Rightarrow Q\leq -(2\sqrt{2}+2)$ hay $Q_{\max}=-(2\sqrt{2}+2)$