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\(P=\sqrt{x^4+x^2y^2}+x^2=\sqrt{x^4+\frac{1}{x^2}}+x^2\)
Ta có: \(x^4+\frac{1}{x^2}=x^4+\frac{1}{8x^2}+\frac{1}{8x^2}+...+\frac{1}{8x^2}\ge9\sqrt[9]{x^4.\left(\frac{1}{8x^2}\right)^8}\)
\(=9\sqrt[9]{\frac{1}{8^8.x^{12}}}\)
=> \(P=3\sqrt[18]{\frac{1}{8^8.x^{12}}}+x^2\)
\(=\sqrt[18]{\frac{1}{8^8x^{12}}}+\sqrt[18]{\frac{1}{8^8x^{12}}}+\sqrt[18]{\frac{1}{8^8x^{12}}}+x^2\)
\(\ge4\sqrt[4]{\left(\sqrt[18]{\frac{1}{8^8x^{12}}}\right)^3.x^2}\)
\(=4.\left(\frac{1}{8^{\frac{1}{3}}.x^{\frac{1}{2}}}\right).x^2=2\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x^4=\frac{1}{8x^2}\\x^2=\sqrt[8]{\frac{1}{8^8x^{12}}}\end{cases}}\)<=> x^2 = 1/2 khi đó y = 2 , x = \(\frac{1}{\sqrt{2}}\)
Vậy GTNN của P = 2.
Từ giả thiết ta có: \(\left(x+y-z\right)^2=4xy\)
\(\Rightarrow P=x+y+z+\frac{2}{\left(x+y-z\right)^2.z}=x+y+z+\frac{8}{4z\left(x+y-z\right)^2}\)
Am-Gm:\(\left(x+y-z\right)\left(x+y-z\right).4z\le\frac{1}{27}\left(2x+2y+2z\right)^3=\frac{8}{27}\left(x+y+z\right)^3\)
\(\Rightarrow P\ge x+y+z+\frac{27}{\left(x+y+z\right)^3}\)
\(=\frac{x+y+z}{3}+\frac{x+y+z}{3}+\frac{x+y+z}{3}+\frac{27}{\left(x+y+z\right)^3}\ge4\sqrt[4]{\frac{\left(x+y+z\right)^3.27}{27.\left(x+y+z\right)^3}}=4\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x+y-z=4z\\x+y+z=3\\\left(x+y-z\right)^2=4xy\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}z=\frac{1}{2}\\x+y=\frac{5}{2}\\xy=1\end{matrix}\right.\)
\(\Rightarrow\left(x;y;z\right)=\left(\frac{1}{2};2;\frac{1}{2}\right)\) hoặc \(\left(2;\frac{1}{2};\frac{1}{2}\right)\). Nhưng vì đề bài cho đối xứng với cả 3 biến nên dấu = xảy ra tại hoán vị của \(\left(2;\frac{1}{2};\frac{1}{2}\right)\)
Vậy P min =4
Ngọc HnueThảo PhươngĐỖ CHÍ DŨNGMinh AnBăng Băng 2k6Vũ Minh Tuấn
B= \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\)
ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=>\(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\frac{9}{16}\)
=> min B=9/16 kh x=-1/2
C= \(x^2-2xy+y^2+1\)= \(\left(x-y\right)^2+1\)
ta có \(\left(x-y\right)^2\ge0\)=>\(\left(x-y\right)^2+1\ge1\)
=> Min C=1 khi x=y
\(1+xy=2\left(x^2+y^2\right)\ge4\left|xy\right|\ge4xy\)
\(\Rightarrow3xy\le1\Rightarrow xy\le\frac{1}{3}\)
\(1+xy\ge4\left|xy\right|\ge-4xy\Rightarrow5xy\ge-1\Rightarrow xy\ge-\frac{1}{5}\)
\(\Rightarrow-\frac{1}{5}\le xy\le\frac{1}{3}\)
\(P=7\left(x^4+y^4+2x^2y^2\right)-10x^2y^2=7\left(x^2+y^2\right)^2-10x^2y^2\)
\(P=\frac{7}{4}\left(xy+1\right)^2-10x^2y^2=-\frac{33}{4}x^2y^2+\frac{7}{2}xy+\frac{7}{4}\)
Đặt \(t=xy\Rightarrow P=f\left(t\right)=-\frac{33}{4}t^2+\frac{7}{2}t+\frac{7}{4}\) với \(t\in\left[-\frac{1}{5};\frac{1}{3}\right]\)
Xét \(f\left(t\right)\) trên \(\left[-\frac{1}{5};\frac{1}{3}\right]\)
\(f\left(-\frac{1}{5}\right)=\frac{18}{25}\) ; \(f\left(\frac{1}{3}\right)=2\) ; \(f\left(-\frac{b}{2a}\right)=f\left(\frac{7}{33}\right)=\frac{70}{33}\)
\(\Rightarrow M=\frac{70}{33}\) ; \(m=\frac{18}{25}\)
Ta có \(xy\le\dfrac{\left(x+y\right)^2}{4}\).
Do đó ta có: \(x+y+xy=x+y-2xy+3xy\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow x^2+y^2\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Leftrightarrow\dfrac{1}{4}\left(x+y\right)^2-\left(x+y\right)\le0\)
\(\Leftrightarrow\left(x+y\right)\left[\dfrac{1}{4}\left(x+y\right)-1\right]\le0\)
\(\Leftrightarrow0\le x+y\le4\).
Do đó m = 0, n = 4.
Vậy m2 + n2 = 16. Chọn A.
Dạ, em cảm ơn