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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.
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
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?
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{x-9}\right)\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{3x+3}{x-9}\right]\) \(\left[\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right]\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}+3x+3}{x-9}.\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(P=\frac{6x-3\sqrt{x}+3}{x-9}.\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\frac{16x}{3-x}+\frac{3}{x}+1=\frac{16x}{3-x}+\frac{3-x}{x}+2\ge8+2=10\)
Dau '=' xay ra khi \(x=\frac{3}{5}\)
Vay \(A_{min}=10\)khi \(x=\frac{3}{5}\)
Ta có : P = \(1-3x+\frac{3}{2-x}=6-3x+\frac{3}{2-x}\) \(-5\) \(=3\left(2-x\right)+\frac{3}{2-x}-5\)
Áp dụng BĐT : AM-GM ta được :
\(3\left(2-x\right)+\frac{3}{2-x}\ge2\sqrt{\frac{3\left(2-x\right)3}{\left(2-x\right)}}=\)\(2\sqrt{9}=2.3=6\)
Vì x<2 => dấu "=" xảy ra khi : x=1
=> P \(\ge6-5=1\)
Vậy Min P = 1 khi x=1