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\(x+\dfrac{16}{x-1}\\ =x-1+\dfrac{16}{x-1}+1\)
Áp dụng BĐT Cô-si ta có:
\(x-1+\dfrac{16}{x-1}+1\\
\ge2\sqrt{\left(x-1\right).\dfrac{16}{x-1}}+1\\
=2\sqrt{16}+1\\
=9\)
Dấu "=" xảy ra
\(\Leftrightarrow x-1=\dfrac{16}{x-1}\\ \Leftrightarrow\left(x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
\(A=2x\left(6-x\right)\le\dfrac{1}{2}\left(x+6-x\right)^2=18\)
Dấu "=" xảy ra khi \(x=3\)
\(B^2=x^2\left(9-x\right)=-x^3+9x^2\)
\(B^2=-x^3+9x^2-108+108=108-\left(x-6\right)^2\left(x+3\right)\le108\)
\(\Leftrightarrow B\le6\sqrt{3}\)
\(C^2=\left(6-x\right)^2x=32-\left(8-x\right)\left(x-2\right)^2\le32\)
\(\Rightarrow C\le4\sqrt{2}\)
\(x^2+y^2\ge2\sqrt{x^2y^2}\ge2xy\)
\(x^2y^2+1\ge2\sqrt{x^2y^2.1}\ge2xy\)
\(\Rightarrow x^2+y^2+x^2.y^2+1\ge2xy+2xy=4xy\)
\(A=x\sqrt{2-x^2}\le\frac{1}{2}\left(x^2+2-x^2\right)=1\)
Dấu "=" xảy ra khi \(x=1\)
\(A=\frac{\sqrt[4]{3}}{2}.\frac{2x}{\sqrt[4]{3}}\sqrt{4-x^4}\le\frac{\sqrt[4]{3}}{4}\left(\frac{4x^2}{\sqrt{3}}+4-x^4\right)=\frac{\sqrt[4]{3}}{4}\left[\frac{16}{3}-\left(x^2-\frac{2\sqrt{3}}{3}\right)^2\right]\le\frac{4\sqrt[4]{3}}{3}\)
\(A_{max}=\frac{4\sqrt[4]{3}}{3}\) khi \(x^2=\frac{2\sqrt{3}}{3}\)
Điều kiện \(a>0\)
\(A=\sqrt[4]{\frac{3}{4a}}.\sqrt[4]{\frac{4a}{3}}.x\sqrt{a-x^4}\le\sqrt[4]{\frac{3}{4a}}\left(-x^4+\sqrt{\frac{4a}{3}}x^2+a\right)\)
\(A\le\sqrt[4]{\frac{3}{4a}}\left[\frac{4a}{3}-\left(x^2-\sqrt{\frac{a}{3}}\right)^2\right]\le\frac{4a}{3}\sqrt[4]{\frac{3}{4a}}\)
Dấu "=" xảy ra khi \(x=\sqrt[4]{\frac{a}{3}}\)