Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A\ge\frac{1}{3}\left(x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}\right)^2\ge\frac{1}{3}\left(x+y+z+\frac{9}{x+y+z}\right)^2=\frac{100}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
\(\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=2+x^2y^2+\frac{1}{x^2y^2}\)
Áp dụng bđt AM-GM và bđt Cauchy-Schwarz:
\(x^2y^2+\frac{1}{x^2y^2}=x^2y^2+\frac{1}{16x^2y^2}+\frac{15}{16x^2y^2}\)
\(\ge2\sqrt{x^2y^2.\frac{1}{16x^2y^2}}+\frac{15}{16x^2y^2}=8+\frac{15}{16x^2y^2}\)
Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow x^2y^2\le\frac{1}{16}\Rightarrow16x^2y^2\le1\Rightarrow\frac{15}{16x^2y^2}\ge15\)
\(\Rightarrow8+\frac{15}{16x^2y^2}\ge23\)
\("="\Leftrightarrow x=y=\frac{1}{2}\)
bạn thay x=1-y rồi thay vào H sau đó làm bình thường nhé
\(I=2+x+\frac{1}{x}+y+\frac{1}{y}+\frac{x}{y}+\frac{y}{x}\)
\(I=2+x+\frac{1}{2x}+y+\frac{1}{2y}+\frac{x}{y}+\frac{y}{x}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(I\ge2+2\sqrt{\frac{x}{2x}}+2\sqrt{\frac{y}{2y}}+2\sqrt{\frac{xy}{xy}}+\frac{1}{2}.\frac{4}{\left(x+y\right)}\)
\(I\ge4+2\sqrt{2}+\frac{2}{x+y}\ge4+2\sqrt{2}+\frac{2}{\sqrt{2\left(x^2+y^2\right)}}=4+3\sqrt{2}\)
\(\Rightarrow I_{min}=4+3\sqrt{2}\) khi \(x=y=\frac{1}{\sqrt{2}}\)
\(K=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
Ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2\ge4\)
\(\Rightarrow\left(y+\frac{1}{y}\right)^2\ge4\)
\(\Rightarrow M\ge8\)
\(K\ge\frac{1}{2}\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2=\frac{1}{2}\left(4x+\frac{1}{x}+4y+\frac{1}{y}-3\left(x+y\right)\right)^2\)
\(K\ge\frac{1}{2}\left(2\sqrt{\frac{4x}{x}}+2\sqrt{\frac{4y}{y}}-3.1\right)^2=\frac{25}{2}\)
\(\Rightarrow K_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)
\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Dấu \("="\Leftrightarrow x=y\)
\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)
\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)
Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)
Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)
Dấu \("="\Leftrightarrow x=y=z=1\)
?Amanda?, Phạm Lan Hương, Phạm Thị Diệu Huyền, Vũ Minh Tuấn, Nguyễn Ngọc Lộc , @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @Trần Thanh Phương
giúp e với ạ! Cần trước 5h chiều nay! Cảm ơn mn nhiều!
Tranh thủ làm 1, 2 bài rồi ăn cơm:
1/ Đặt \(m=n-2008>0\)
\(\Rightarrow2^{2008}\left(369+2^m\right)\) là số chính phương
\(\Rightarrow369+2^m\) là số chính phương
m lẻ thì số trên chia 3 dư 2 nên ko là số chính phương
\(\Rightarrow m=2k\Rightarrow369=x^2-\left(2^k\right)^2=\left(x-2^k\right)\left(x+2^k\right)\)
b/
\(2\left(a^2+b^2\right)\left(a+b-2\right)=a^4+b^4\) \(\left(a+b>2\right)\)
\(\Rightarrow2\left(a^2+b^2\right)\left(a+b-2\right)\ge\frac{1}{2}\left(a^2+b^2\right)^2\)
\(\Rightarrow a^2+b^2\le4\left(a+b-2\right)\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2\le0\Rightarrow a=b=2\)
\(\Rightarrow x=y=4\)
M= \(x^2y^2+2+\frac{1}{x^2y^2}=\left(xy+\frac{1}{xy}\right)^2\)
\(xy+\frac{1}{xy}=xy+\frac{1}{16xy}+\frac{15}{16xy}\ge2\sqrt{xy.\frac{1}{16xy}}+\frac{15\left(x+y\right)}{16xy}=\frac{1}{2}+\frac{15}{16}\left(\frac{1}{x}+\frac{1}{y}\right)\ge\)\(\frac{1}{2}+\frac{15}{16}.\frac{4}{x+y}=\frac{1}{2}+\frac{15}{16}.4=\frac{17}{4}\) => M\(\ge\frac{17^2}{4^2}\)
dấu '=' khi xy = \(\frac{1}{16xy};x=y=>x=y=\frac{1}{2}\)
\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=\frac{x^2y^2+1}{y^2}.\frac{y^2x^2+1}{x^2}=\frac{\left(x^2y^2+1\right)^2}{x^2y^2}\)
\(=\frac{x^4y^4+2x^2y^2+1}{x^2y^2}=x^2y^2+2+\frac{1}{x^2y^2}=\left(xy+\frac{1}{xy}\right)^2\)
ta có:\(xy+\frac{1}{xy}=16xy+\frac{1}{xy}-15xy \left(1\right) \)
mặt khác:\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow-15xy\ge-\frac{15}{4} \left(2\right)\)
áp dụng bất đẳng thức cô si ta có:\(16xy+\frac{1}{xy}\ge2\sqrt{16xy.\frac{1}{xy}}=8 \left(3\right)\)
từ (1), (2), (3) ta có\(xy+\frac{1}{xy}\ge8-\frac{15}{4}=\frac{17}{4}\Rightarrow\left(xy+\frac{1}{xy}\right)^2\ge\frac{289}{16}\)
vậy \(M_{min}=\frac{289}{16}\)đạt được khi \(x=y=\frac{1}{2}\)