Dự đoán dấu "=" khi x = 2 ; y= 1

Áp dụng bđt Cô-si cho 3 số và bđt \(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\) ta được

\(P=2x^2+y^2+\frac{28}{x}+\frac{1}{y}\)

    \(=\left(\frac{7x^2}{4}+\frac{14}{x}+\frac{14}{x}\right)+\left(\frac{y^2}{2}+\frac{1}{2y}+\frac{1}{2y}\right)+\left(\frac{x^2}{4}+\frac{y^2}{2}\right)\)

    \(\ge3\sqrt[3]{\frac{7x^2.14.14}{4.x^2}}+3\sqrt[3]{\frac{y^2.1.1}{2.2y.2y}}+\frac{\left(x+y\right)^2}{4+2}\)

      \(=3.\sqrt[3]{\frac{7.14.14}{4}}+\frac{3}{\sqrt[3]{2^3}}+\frac{3^2}{6}=24\)

Dấu "=" khi x = 2 ; y = 1 

Bài toán easy!

\(P=\left(2x^2+8\right)+\left(y^2+1\right)+\frac{28}{x}+\frac{1}{y}-9\)

Áp dụng BĐT AM-GM,ta có:

\(P\ge8x+2y+\frac{28}{x}+\frac{1}{y}-9\)

\(=\left(7x+\frac{28}{x}\right)+\left(y+\frac{1}{y}\right)+\left(x+y\right)-9\)

\(\ge2\sqrt{7x.\frac{28}{x}}+2\sqrt{y.\frac{1}{y}}+\left(x+y\right)-9\)

\(\ge28+2+3-9=24\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x^2=8\\y^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Vậy \(P_{min}=24\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Làm xong rồi mới biết anh Incursion đã làm rồi -_-.Nãy giờ cứ chăm chú vào máy tính casio để tìm dấu "=" nên chả để ý=((

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).

Đẳng thức xảy ra khi x = 1; y = 2.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

Ta có: \(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra <=> \(\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra <=> 2x=y và xy=2

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra <=> x=1, y=2

Vậy GTNN của M là 11/4 <=> x=1;y=2

Ta có:

\(M=\dfrac{2x+y}{xx}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow2x=y,xy=2\)

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow x=1,y=2\)

Vậy GTNN của M là \(\dfrac{11}{4}\Leftrightarrow x=1,y=2\)

sai đè nha:4\(\sqrt{yz}\)

cây gì lớn nhất hành tinh

\(y=2+\dfrac{6}{x-3}\)

\(P=3x\left(2+\dfrac{6}{x-3}\right)+2x+2+\dfrac{6}{x-3}\)

\(P=8x+2+\dfrac{18x}{x-3}+\dfrac{6}{x-3}=8x+20+\dfrac{60}{x-3}\)

\(P=8\left(x-3\right)+\dfrac{60}{x-3}+44\ge2\sqrt{\dfrac{480\left(x-3\right)}{x-3}}+44=44+8\sqrt{30}\)

\(P_{min}=44+8\sqrt{30}\) khi \(8\left(x-3\right)=\dfrac{60}{x-3}\Leftrightarrow x=\dfrac{6+\sqrt{30}}{2}\)

Nếu tồn tại 1 số bằng 0 \(\Rightarrow P=1\)

Nếu x;y đều dương:

\(P=\dfrac{x^2}{xy+x}+\dfrac{y^2}{xy+y}\ge\dfrac{\left(x+y\right)^2}{2xy+x+y}\ge\dfrac{\left(x+y\right)^2}{\dfrac{1}{2}\left(x+y\right)^2+x+y}=\dfrac{2}{3}\)

\(P_{min}=\dfrac{2}{3}\) khi \(x=y=\dfrac{1}{2}\)

Bài này có thể tìm được cả max:

\(\left\{{}\begin{matrix}y+1\ge1\Rightarrow\dfrac{x}{y+1}\le x\\x+1\ge1\Rightarrow\dfrac{y}{x+1}\le y\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{x}{y+1}+\dfrac{y}{x+1}\le x+y=1\)

\(P_{max}=1\) khi \(\left(x;y\right)=\left(0;1\right)\) và hoán vị

Ta có: 3x + y = 1 => y = 1 - 3x

a, Thay y = 1 - 3x vào M, ta có:

\(\Rightarrow M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=3\left(4x^2-2x+\frac{1}{3}\right)\)

\(=3\left(4x^2-2x+\frac{1}{4}+\frac{1}{12}\right)=3\left(2x-\frac{1}{2}\right)^2+\frac{3}{12}=3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-\frac{1}{2}=0\\3x+y=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=1-3x=1-3.\frac{1}{4}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{4}\)

Vậy GTNN M = 1/4 khi x = y = 1/4

b, Thay y = 1 - 3x vào N

\(\Rightarrow N=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{x}{3}+\frac{1}{36}-\frac{1}{36}\right)\)

\(=-3\left(x-\frac{1}{6}\right)^2-3.\left(-\frac{1}{36}\right)=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)

Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2\le0\forall x\)

\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{6}=0\\3x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=1-3x=1-3.\frac{1}{6}=\frac{1}{2}\end{cases}}\)

Vậy GTLN N = 1/12 khi x = 1/6 và y = 1/2