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Ta có:
\(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)
\(\Leftrightarrow xy\ge1\)
Theo đề bài thì
\(\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{y^4+x^2+2yx^2}\le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}\le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
Câu hỏi của Anh Tú Dương - Toán lớp 10 | Học trực tuyến
\(\left(x^2+\dfrac{8}{27x}+\dfrac{8}{27x}\right)+\left(y^2+\dfrac{8}{27y}+\dfrac{8}{27y}\right)+\dfrac{11}{27}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\ge3\sqrt[3]{\dfrac{8^2}{27^2}}+3\sqrt[3]{\dfrac{8^2}{27^2}}+\dfrac{11}{27}.\dfrac{4}{x+y}\)
\(\ge\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{11}{9}=\dfrac{35}{9}\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
1) \(y=\dfrac{2x^2+1}{x^3-5x+4}\)
ĐK \(x^3-5x+4\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\dfrac{\sqrt{17}-1}{2}\\x\ne\dfrac{-\sqrt{17}-1}{2}\end{matrix}\right.\)
TXĐ \(D=R\backslash\left\{1;\dfrac{\sqrt{17}-1}{2};\dfrac{-\sqrt{17}-1}{2}\right\}\)
2) \(y=\dfrac{\sqrt{x-2}}{\left(x-3\right)^3-1}\)
ĐK \(\left\{{}\begin{matrix}x-2\ge0\\x-3\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne4\end{matrix}\right.\)
TXĐ \(D=[2;+\infty)\backslash\left\{4\right\}\)
3) \(y=\sqrt{x-2}-\dfrac{2}{\sqrt[3]{x-1}}\)
ĐK\(\left\{{}\begin{matrix}x+2\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\ne1\end{matrix}\right.\)
TXĐ \(D=[-2;+\infty)\backslash\left\{1\right\}\)
4) \(y=\dfrac{x^2+2}{\sqrt{\left(x+3\right)^2}}=\dfrac{x^2+2}{\left|x-3\right|}\)
ĐK \(x-3\ne0\Leftrightarrow x\ne3\)
TXĐ \(D=R\backslash\left\{3\right\}\)
5) \(y=\dfrac{\sqrt{x^2-2}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
ĐK \(\left\{{}\begin{matrix}x^2-2\ge0\\x>0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in(-\infty;-\sqrt{2}]\cap[\sqrt{2};+\infty)\\x>0\\x\ne9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge\sqrt{2}\\x\ne9\end{matrix}\right.\)
TXĐ \(D=[\sqrt{2};+\infty)\backslash\left\{9\right\}\)
6) \(y=\sqrt{1-\sqrt{1+x}}\)
ĐK \(\left\{{}\begin{matrix}x+1\ge0\\1-\sqrt{1+x}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1\ge\sqrt{1+x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1\ge1+x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le0\end{matrix}\right.\)
TXĐ \(D=\left[0;-1\right]\)
Ta có \(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Leftrightarrow xy\ge1\)
\(A=\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{x^2+y^4+2x^2y}\\ \le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}=\dfrac{1}{4xy}+\dfrac{1}{4xy}\\ \le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
Dấu \("="\Leftrightarrow x=y=1\)