Áp dụng BĐT AM-GM ta có:
\(1=\frac{3}{x}+\frac{2}{y}\ge2.\sqrt{\frac{6}{xy}}\)
\(\Leftrightarrow1^2\ge4.\frac{6}{xy}\)
\(\Leftrightarrow1\ge\frac{24}{xy}\)
\(\Leftrightarrow xy\ge24\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{3}{x}=\frac{2}{y}=\frac{1}{2}\Leftrightarrow\hept{\begin{cases}x=6\\y=4\end{cases}}\)
Vậy \(xy_{min}=24\Leftrightarrow\hept{\begin{cases}x=6\\y=4\end{cases}}\)
T nghĩ ra câu b rồi nhé Pain,bớt xạo lz!
b) Từ \(\frac{3}{x}+\frac{2}{y}=1\),ta có: \(x+y=1\left(x+y\right)=\left(\frac{3}{x}+\frac{2}{y}\right)\left(x+y\right)\)
Áp dụng BĐT Bunhiacopxki,ta có: \(\left(\frac{3}{x}+\frac{2}{y}\right)\left(x+y\right)\ge\left(\sqrt{\frac{3}{x}.x}+\sqrt{\frac{2}{y}.y}\right)\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2=5+2\sqrt{6}\)
Vậy \(Min_{x+y}=5+2\sqrt{6}\Leftrightarrow\hept{\begin{cases}x=3+\sqrt{6}\\y=2+\sqrt{6}\end{cases}}\)
\(1=\frac{3}{x}+\frac{2}{y}=\frac{3y+2x}{xy}\)\(\Leftrightarrow\)\(xy=2x+3y\)
\(1=\frac{3}{x}+\frac{2}{y}=\frac{6}{2x}+\frac{6}{3y}=6\left(\frac{1}{2x}+\frac{1}{3y}\right)\ge\frac{6\left(1+1\right)^2}{2x+3y}=\frac{24}{2x+3y}\)
\(\Leftrightarrow\)\(2x+3y\ge24\)
\(\Rightarrow\)\(xy=2x+3y\ge24\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{1}{2x}=\frac{1}{3y}\)\(\Leftrightarrow\)\(x=\frac{3y}{2}\)
\(\frac{3}{x}+\frac{2}{y}=1\)\(\Leftrightarrow\)\(\frac{3}{\frac{3y}{2}}+\frac{2}{y}=1\)\(\Leftrightarrow\)\(y=4\)\(\Rightarrow\)\(x=\frac{3y}{2}=\frac{3.4}{2}=6\)
Tôi chỉ làm được câu a).Câu b) cần người giúp:
a) Bình phương hai vế của giả thiết:\(\left(\frac{3}{x}+\frac{2}{y}\right)^2=1\Leftrightarrow\frac{9}{x^2}+\frac{12}{xy}+\frac{4}{y^2}=1\)
Áp dụng BĐT Cô si,ta có:
\(1=\left(\frac{9}{x^2}+\frac{4}{y^2}\right)+\frac{12}{xy}\ge2\sqrt{\frac{9.4}{x^2y^2}}+\frac{12}{xy}\)
\(=\frac{12}{xy}+\frac{12}{xy}=\frac{24}{xy}\)
Ta có: \(1\ge\frac{24}{xy}\Rightarrow xy\ge24\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{9}{x^2}=\frac{4}{y^2}\\\frac{3}{x}+\frac{2}{y}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{3}{x}=\frac{2}{y}\\\frac{3}{x}+\frac{2}{y}=1\end{cases}}\)
\(\Leftrightarrow\frac{3}{x}=\frac{2}{y}=\frac{1}{2}\Leftrightarrow\hept{\begin{cases}x=6\\y=4\end{cases}}\)
\(3y+2x=1\)
\(2\left(x+y\right)=1-y\)
\(4\left(x+y\right)^2=\left(1-y\right)^2\)
\(4\left(x+y\right)^2=1^2-2y+y^2\)
\(\left(x+y\right)^2=\frac{\left(1+y^2\right)-2y}{4}\ge\frac{2y-2y}{4}=0\)
\(\left(\sqrt{\frac{3y}{x}}-\sqrt{\frac{2x}{y}}\right)^2\ge0\)
\(\frac{3y}{x}+\frac{2x}{y}\ge12\)
\(\left(\frac{3y}{x}+3\right)+\left(\frac{2x}{y}+2\right)\ge3+12+2\)
\(\frac{3}{x}\left(y+x\right)+\frac{2}{y}\left(x+y\right)\ge17\)
\(\left(x+y\right)\left(\frac{3}{x}+\frac{2}{y}\right)\ge17\)
\(\left(x+y\right)\ge17\)
\(\left(\sqrt{\frac{3y}{x}}+\sqrt{\frac{2x}{y}}\right)^2\ge0.\)
\(\frac{3y}{x}+\frac{2x}{y}+5\ge2\sqrt{6}+5\)
\(\left(\frac{3y}{x}+3\right)+\left(\frac{2x}{y}+2\right)\ge2\sqrt{6}+5\)
\(\frac{3}{x}\left(y+x\right)+\frac{2}{y}\left(y+x\right)\ge2\sqrt{6}+5\)
\(\left(x+y\right).1\ge....\)
sửa lại dòng đầu thành
\(\left(\sqrt{\frac{3y}{x}}-\sqrt{\frac{2x}{y}}\right)^2\)
ối, có người tự ra câu hỏi, tự trả lời
