cho các số thực x,y thỏa mãn x2+y2=2017. tìm min và max của bt P=xy+x+y
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\(\left(x^2+9\right)+\left(y^2+9\right)+3\left(x^2+y^2\right)\ge6x+6y+6xy=90\)
\(\Rightarrow4\left(x^2+y^2\right)+18\ge90\)
\(\Rightarrow x^2+y^2\ge18\)
\(P_{min}=18\) khi \(x=y=3\)
\(x+y+xy=15\Rightarrow\left\{{}\begin{matrix}x\le15\\y\le15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\left(x-15\right)\le0\\y\left(y-15\right)\le0\end{matrix}\right.\)
\(\Rightarrow x^2+y^2\le15x+15y\) (1)
Cũng từ đó ta có: \(\left(x-15\right)\left(y-15\right)\ge0\Rightarrow xy\ge15x+15y-225\)
\(\Rightarrow16x+16y-225\le x+y+xy=15\)
\(\Rightarrow x+y\le15\) (2)
(1);(2) \(\Rightarrow x^2+y^2\le15.15=225\)
\(P_{max}=225\) khi \(\left(x;y\right)=\left(0;15\right);\left(15;0\right)\)
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\(\dfrac{x^2+y^2}{2}\ge xy\Rightarrow-xy\ge-\dfrac{x^2+y^2}{2}\)
\(\Rightarrow4=x^2+y^2-xy\ge x^2+y^2-\dfrac{x^2+y^2}{2}=\dfrac{x^2+y^2}{2}\)
\(\Rightarrow x^2+y^2\le8\)
\(C_{max}=8\) khi \(x=y=\pm2\)
\(x^2+y^2\ge-2xy\Rightarrow-xy\le\dfrac{x^2+y^2}{2}\)
\(4=x^2+y^2-xy\le x^2+y^2+\dfrac{x^2+y^2}{2}=\dfrac{3}{2}\left(x^2+y^2\right)\)
\(\Rightarrow x^2+y^2\ge\dfrac{8}{3}\)
\(C_{min}=\dfrac{8}{3}\) khi \(\left(x;y\right)=\left(-\dfrac{2}{\sqrt{3}};\dfrac{2}{\sqrt{3}}\right);\left(\dfrac{2}{\sqrt{3}};-\dfrac{2}{\sqrt{3}}\right)\)
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Do \(x^2+y^2=1\Rightarrow-1\le x;y\le1\Rightarrow\left\{{}\begin{matrix}y+1\ge0\\1-y\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y^2\left(y+1\right)\ge0\\y^2\left(1-y\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y^3\ge-y^2\\y^3\le y^2\end{matrix}\right.\)
Với mọi số thực x ta có:
\(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x\ge-x^2-1\\2x\le x^2+1\end{matrix}\right.\)
Do đó: \(\left\{{}\begin{matrix}P=2x+y^3\ge-x^2-1-y^2=-2\\P=2x+y^3\le x^2+1+y^2=2\end{matrix}\right.\)
\(P_{min}=-2\) khi \(\left(x;y\right)=\left(-1;0\right)\)
\(P_{max}=2\) khi \(\left(x;y\right)=\left(1;0\right)\)
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\(A=x-2y+3\Rightarrow x=A+2y-3\)
\(\Rightarrow\left(2y+A-3\right)^2+y\left(A+2y-3\right)+2y^2=1\)
\(\Leftrightarrow8y^2+\left(5A-15\right)y+A^2-6A+8=0\)
\(\Delta=\left(5A-15\right)^2-32\left(A^2-6A+8\right)\ge0\)
\(\Leftrightarrow-7A^2+42A-31\ge0\)
\(\Rightarrow\dfrac{21-4\sqrt{14}}{7}\le A\le\dfrac{21+4\sqrt{14}}{7}\)
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Từ đề bài \(\Rightarrow4x^2+4y^2+4xy-24x-24y+44=0\)
\(\Leftrightarrow\left(2x+y\right)^2-24x-12y+36+3y^2-12y+12-4=0\)
\(\Leftrightarrow\left(2x+y-6\right)^2+3\left(y-2\right)^2-4=0\)
\(\Leftrightarrow\left(2x+y-6\right)^2=4-3\left(y-2\right)^2\le4\forall x;y\)
\(\Leftrightarrow-2\le2x+y-6\le2\Rightarrow4\le2x+y\le8\)
Do đó \(4\le P\le8\)
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1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)