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1,\(A=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\)\(\ge2\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=\frac{3}{2}\)
Vậy Amin\(=\frac{3}{2}\Leftrightarrow\frac{x-1}{2}=\frac{2}{x-1}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Xét ĐK ta thấy x=3.
2,Áp dụng bđt Cô-si:
...........\(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}\ge2y^2\)
...........\(\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2}\ge2z^2\)
\(\frac{x^2z^2}{y^2}+\frac{x^2y^2}{z^2}\ge2x^2\)
Mk nghĩ đề phải là x^2+y^2+z^2=1
\(\Rightarrow VT\ge x^2+y^2+z^2=1\)
Vậy Amin=1 khi \(x=y=z=\sqrt{\frac{1}{3}}=\frac{\sqrt{3}}{3}\)
Câu cuối chưa bt làm.
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{x-9}\right)\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{3x+3}{x-9}\right]\) \(\left[\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right]\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}+3x+3}{x-9}.\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(P=\frac{6x-3\sqrt{x}+3}{x-9}.\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(x^2-5x+6\ge0\)
\(x^2-2x-3x+6\ge0\)
\(x\left(x-2\right)-3\left(x-2\right)\ge0\)
\(\left(x-3\right)\left(x-2\right)\ge0\)
\(\Rightarrow\)\(\hept{\begin{cases}x-3\ge0\\x-2\ge0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3\le0\\x-2\le0\end{cases}}\)
\(\Rightarrow\)\(\hept{\begin{cases}x\ge3\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le3\\x\le2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\ge3\\x\le2\end{cases}}\)
vậy tập nghiệm của phương trình là \(\orbr{\begin{cases}x\ge3\\x\le2\end{cases}}\)
\(x^2-6x+8< 8\)
\(x^2-4x-2x+8< 0\)
\(x\left(x-4\right)-2\left(x-4\right)< 0\)
\(\left(x-2\right)\left(x-4\right)< 0\)
\(\Rightarrow\)\(\hept{\begin{cases}x-2>0\\x-4< 0\end{cases}}\) hoặc \(\hept{\begin{cases}x-2< 0\\x-4>0\end{cases}}\)
\(\Rightarrow\)\(\hept{\begin{cases}x>2\\x< 4\end{cases}}\) hoặc \(\hept{\begin{cases}x< 2\\x>4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2< x< 4\\\varnothing\end{cases}}\)
vậy \(2< x< 4\) hay \(x=3\)
\(\frac{x-1}{3}-\frac{2x+1}{2}< \frac{5x+1}{6}-x\)
\(\frac{\left(x-1\right).2}{6}-\frac{\left(2x+1\right).3}{6}< \frac{5x+1}{6}-\frac{6x}{6}\)
\(2x-2-6x-3< 5x+1-6x\)
\(-3x< 6\)
\(x>-2\)
vậy tập nghiệm của bất phương trình là \(x>-2\)
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
Ta có:
\(A=\left(x^2+\frac{1}{8x}+\frac{1}{8x}\right)+\left(y^2+\frac{1}{8y}+\frac{1}{8y}\right)+\left(z^2+\frac{1}{8z}+\frac{1}{8z}\right)+\frac{6}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}+3\sqrt[3]{y^2.\frac{1}{8y}.\frac{1}{8y}}+3\sqrt[3]{z^2.\frac{1}{8z}.\frac{1}{8z}}+\frac{6}{8}\frac{9}{x+y+z}\)
\(=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{6}{8}.\frac{9}{\frac{3}{2}}=\frac{27}{4}\)
Dấu "=" xảy ra <=> x = y = z = 1/2
Vậy min A = 27/4 tại x = y = z = 1/2
\(P=\frac{1}{x}+\frac{2}{1-2x}=2\left(\frac{1}{2x}+\frac{1}{1-2x}\right)\ge2.\frac{4}{2x+1-2x}=8\)
Dau '=' xay ra khi \(x=\frac{1}{4}\)
Vay \(P_{min}=8\)khi \(x=\frac{1}{4}\)