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a/ \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}.\frac{18}{x}}=...\)
b/ \(\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}.\frac{2}{x-1}}+\frac{1}{2}=...\)
c/ \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=...\)
d/ \(\frac{x}{3}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=...\)
e/ \(\frac{x}{1-x}+\frac{5}{x}=\frac{x}{1-x}+\frac{5-5x+5x}{x}=\frac{x}{1-x}+\frac{5\left(1-x\right)}{x}+5\ge2\sqrt{\frac{x}{1-x}.\frac{5\left(1-x\right)}{x}}+5=...\)
f/ \(\frac{x^3+1}{x^2}=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge2\sqrt{\frac{x}{2}.\frac{x}{2}.\frac{1}{x^2}}=...\)
g/ \(\frac{x^2+4x+4}{x}=x+\frac{4}{x}+4\ge2\sqrt{x.\frac{4}{x}}+4=...\)
Mình áp dụng luôn Cô - si cho các số ta được
a) \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}\cdot\frac{18}{x}}=2.\sqrt{9}=2.3=6\)
b) \(y=\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}\cdot\frac{2}{x-1}}+\frac{1}{2}=2+\frac{1}{2}=\frac{5}{2}\)
c) \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}\cdot\frac{1}{x+1}}-\frac{3}{2}=2\sqrt{\frac{3}{2}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)
h) \(x^2+\frac{2}{x^2}\ge2\sqrt{x^2\cdot\frac{2}{x^2}}=2\sqrt{2}\)
g) \(\frac{x^2+4x+4}{x}=\frac{\left(x+2\right)^2}{x}\ge0\)
\(y=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{2\left(x-1\right)}{2\left(x-1\right)}}+\frac{1}{2}=\frac{5}{2}\)
Dấu "=" xảy ra khi \(\frac{x-1}{2}=\frac{2}{x-1}\Rightarrow x=3\)
\(y=\frac{5\left(3x-1\right)}{9}+\frac{5}{3x-1}+\frac{5}{9}\ge2\sqrt{\frac{25\left(3x-1\right)}{9\left(3x-1\right)}}+\frac{5}{9}=\frac{35}{9}\)
Dấu "=" xảy ra khi \(x=\frac{4}{3}\)
\(y=-2+\frac{2}{1-x}+\frac{3}{x}\ge-2+\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{1-x+x}=3+2\sqrt{6}\)
Dấu "=" xảy ra khi \(\frac{1-x}{\sqrt{2}}=\frac{x}{\sqrt{3}}\Rightarrow x=3-\sqrt{6}\)
\(y=x+\frac{9}{x}+2020\ge2\sqrt{\frac{9x}{x}}+2020=2026\)
Dấu "=" xảy ra khi \(x=3\)
a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)
b/
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)
\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)
\(\Leftrightarrow\left|x+2\right|-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
c/
\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)
\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)
Đặt \(\frac{\left|x-2\right|}{x-1}=a\)
\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)
e/ ĐKXĐ: ...
Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)
\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)