Giải chi tiết pt: \(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
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\(Q=\frac{2010,2011}{2014,2015}+\frac{2012,2013}{2014,2015}.\frac{1}{x^2}-\frac{2}{2014,2015}.\frac{1}{x}\)
\(=\frac{2010,2011}{2014,2015}+\frac{2012,2013}{2014,2015}\left(\frac{1}{x^2}-2.\frac{1}{x}.\frac{1}{2012,2013}+\frac{1}{\left(2012,2013\right)^2}\right)-\frac{1}{2012,2013}.\frac{1}{2014,2015}\)
\(=\frac{2010,2011}{2014,2015}-\frac{1}{2012,2013}.\frac{1}{2014,2015}+\frac{2012,2013}{2014,2015}\left(\frac{1}{x}-\frac{1}{2012,2013}\right)^2\)
\(\ge\frac{2010,2011}{2014,2015}-\frac{1}{2012,2013}.\frac{1}{2014,2015}\)
Dấu "=" xảy ra khi và chỉ khi \(x=2012,2013\)
Vậy GTNN của biểu thức là .....
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ĐKXĐ:
\(x\ge0\text{ và }\sqrt{x}-1\ge0\text{ và }x+1\ne0\)
\(\Leftrightarrow x\ge1\text{ và }x\ne-1\)
\(\left(\frac{x\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right):\frac{\sqrt{x}-1}{x+1}\)
\(=\left(\frac{x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}+1}\)
\(=\left(\frac{x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right).\frac{x+1}{\sqrt{x}+1}\)
\(=\frac{x\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\frac{x+1}{\sqrt{x}+1}\)
\(=\frac{x\sqrt{x}+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
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375=(33)25=2725
450=(42)25=1625
vì 2725>1625=>375>450
vậy 375>450
Ta co :375 va 450
=> 375 = (315)5=14348907
=>450 = (410)5 = 1048576
Vi : 14348907 > 1048576
Vay 375 > 450
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