\(xy=1\Rightarrow y=\frac{1}{x}\)
\(M=f\left(x\right)=\frac{x^3}{1+\frac{1}{x}}+\frac{\left(\frac{1}{x}\right)^3}{1+x}=\frac{x^4}{x+1}+\frac{1}{x^3\left(x+1\right)}=\frac{x^7+1}{x^4+x^3}\)
\(f'\left(x\right)=\frac{7x^6\left(x^4+x^3\right)-\left(4x^3+3x^2\right).\left(x^7+1\right)}{\left(x^4+x^3\right)^2}=\frac{3x^{10}+4x^9-4x^3-3x^2}{\left(x^4+x^3\right)^2}=\frac{3x^2\left(x^8-1\right)+4x^3\left(x^6-1\right)}{\left(x^4+x^3\right)^2}\)
\(f'\left(x\right)=0\Rightarrow x=1\)
Dựa vào BBT ta thấy \(f\left(x\right)_{min}=f\left(1\right)=1\)
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