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Ta có \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)=x^2y^2+1+1+\dfrac{1}{x^2y^2}=x^2y^2+2+\dfrac{1}{x^2y^2}=\dfrac{x^4y^4+2x^2y^2+1}{x^2y^2}=\dfrac{\left(x^2y^2+1\right)^2}{\left(xy\right)^2}=\left(\dfrac{x^2y^2+1}{xy}\right)^2=\left(xy+\dfrac{1}{xy}\right)^2=\left(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\right)^2\)
Áp dụng bđt cosi, ta có \(xy+\dfrac{1}{16xy}\ge2\sqrt{xy.\dfrac{1}{16xy}}=2\sqrt{\dfrac{1}{16}}=2.\dfrac{1}{4}=\dfrac{1}{2}\)
\(2\sqrt{xy}\le\left(x+y\right)^2\Leftrightarrow\sqrt{xy}\le\dfrac{\left(x+y\right)^2}{2}=\dfrac{1}{2}\Leftrightarrow xy\le\dfrac{1}{4}\Leftrightarrow\dfrac{15}{16xy}\ge\dfrac{15}{4}\)
Vậy \(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\Leftrightarrow\left(xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\right)^2\ge\dfrac{289}{16}\)
Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x+y=1\\xy=\dfrac{1}{16xy}\\x=y\end{matrix}\right.\)\(\Leftrightarrow\)\(x=y=0,5\)
Vậy GTNN của \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)\)=\(\dfrac{289}{16}\) và xảy ra khi x=y=0,5
\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{2xy}+\frac{\left(x+y\right)^2}{2xy}\)
\(S\ge\frac{4\left(x+y\right)^2}{x^2+y^2+2xy}+\frac{\left(x+y\right)^2}{\frac{\left(x+y\right)^2}{2}}=\frac{4\left(x+y\right)^2}{\left(x+y\right)^2}+2=6\)
\(\Rightarrow S_{min}=6\) khi \(x=y\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) thì bài toán trở thành
Cho \(a+b+ab=3\)
Tìm GTLN của: \(M=\dfrac{3b}{a+1}+\dfrac{3a}{b+1}-a^2-b^2=\dfrac{ab}{a+1}+\dfrac{ab}{b+1}\)
Ta có: \(3=a+b+ab\ge3\sqrt[3]{a^2b^2}\)
\(\Leftrightarrow ab\le1\)
Ta lại có: \(M=\dfrac{ab}{a+1}+\dfrac{ab}{b+1}=ab.\dfrac{a+1+b+1}{ab+a+b+1}=ab.\dfrac{5-ab}{4}\)
\(=\dfrac{5ab-a^2b^2}{4}=\dfrac{\left(-a^2b^2+2ab-1\right)+3ab+1}{4}=\dfrac{-\left(ab-1\right)^2+3ab+1}{4}\le\dfrac{3+1}{4}=1\)
Vậy GTLN là \(M=1\) khi \(a=b=1\) hay \(x=y=1\)
Câu a :
\(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2+x^2+y^2=9\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
Đặt \(x+y=S\) ; \(xy=P\) , phương trình trở thành :
\(\left\{{}\begin{matrix}S^2-2P+P^2=9\\S\left(P-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{3}{P-1}\right)^2-2P+P^2=9\\S=\dfrac{3}{P-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}P=0\\P=-2\\P=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}S=-3\\S=-1\\S=3\end{matrix}\right.\)
Với \(S=-3\) và \(P=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\xy=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\end{matrix}\right.\)
Với \(S=-1\) và \(P=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\xy=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\end{matrix}\right.\)
Với \(S=3\) và \(P=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy phương trình có các cặp nghiệm là : \(\left(x;y\right)=\left(0;-3\right)\) ; \(\left(x;y\right)=\left(-3;0\right)\) ; \(\left(x;y\right)=\left(1;-2\right)\) ; \(\left(x;y\right)=\left(-2;1\right)\) ; \(\left(x;y\right)=\left(2;1\right)\) ; \(\left(x;y\right)=\left(1;2\right)\)
Wish you study well !!
Phùng Khánh Linh Ko đúng đâu ! Bạn thay \(x=y=\dfrac{1}{2}\) vào thì ra tới 10 lận . \(\dfrac{1}{\dfrac{1}{2}}+\dfrac{4}{\dfrac{1}{2}}=10\) lận cơ ?
A=\(1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{y}{x}\)
A= \(\left(x+\dfrac{1}{2x}\right)+\left(y+\dfrac{1}{2y}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+2\)
Áp Dụng BĐT Cô si ta có:
\(\left(x+\dfrac{1}{2x}\right)\ge\sqrt{2}\); \(\left(y+\dfrac{1}{2y}\right)\ge\sqrt{2}\); \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)
\(\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge2\sqrt{\dfrac{1}{2x.2y}}=\dfrac{1}{\sqrt{xy}}\ge\dfrac{\sqrt{2}}{\sqrt{x^2+y^2}}=\sqrt{2}\)
suy ra A\(\ge4+3\sqrt{2}\)
Dấu = xảy ra
\(\left\{{}\begin{matrix}x=y\\x=\dfrac{1}{2x}\\y=\dfrac{1}{2y}\end{matrix}\right.\)
\(\Leftrightarrow\)x=y=\(\dfrac{\sqrt{2}}{2}\)
Vậy Min A=4+3\(\sqrt{2}\) khi x=y=\(\dfrac{\sqrt{2}}{2}\)
Trước hết ta có \(\dfrac{\left(x+y\right)^2}{2}\le x^2+y^2\Rightarrow x+y\le\sqrt{2\left(x^2+y^2\right)}=\sqrt{2}\)
\(A=1+\dfrac{1}{y}+x+\dfrac{x}{y}+1+\dfrac{1}{x}+y+\dfrac{y}{x}\)
\(A=2+x+y+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{x}{y}+\dfrac{y}{x}\ge2+x+y+\dfrac{4}{x+y}+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}\)
\(\Rightarrow A\ge4+x+y+\dfrac{4}{x+y}=4+x+y+\dfrac{2}{x+y}+\dfrac{2}{x+y}\)
\(\Rightarrow A\ge4+2\sqrt{\left(x+y\right).\dfrac{2}{\left(x+y\right)}}+\dfrac{2}{\sqrt{2}}=4+3\sqrt{2}\)
\(\Rightarrow A_{min}=4+3\sqrt{2}\) khi \(x=y=\dfrac{1}{\sqrt{2}}\)
Áp dụng BĐT AM-GM ta có:
\(M=\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)\)
\(=\dfrac{x^2y^2+1}{y^2}\cdot\dfrac{x^2y^2+1}{x^2}=\dfrac{x^4y^4+2x^2y^2+1}{x^2y^2}\)
\(=x^2y^2+\dfrac{1}{x^2y^2}+2=x^2y^2+\dfrac{1}{256x^2y^2}+\dfrac{255}{256x^2y^2}+2\)
\(\ge2\sqrt{x^2y^2\cdot\dfrac{1}{256x^2y^2}}+\dfrac{255}{256\cdot\left(xy\right)^2}+2\)
\(\ge2\cdot\dfrac{1}{16}+\dfrac{255}{256\cdot\left(\dfrac{\left(x+y\right)^2}{4}\right)^2}+2\)
\(=\dfrac{1}{8}+\dfrac{255}{256\cdot\left(\dfrac{1}{4}\right)^2}+2=\dfrac{289}{16}\)
Khi \(x=y=\dfrac{1}{2}\)