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Ta có \(B\ge\dfrac{\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2}{2}\) \(=\dfrac{\left(1+\dfrac{1}{xy}\right)^2}{2}\)
Lại có \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)
\(\Rightarrow B\ge\dfrac{\left(1+4\right)^2}{2}=\dfrac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Vậy GTNN của B là \(\dfrac{25}{2}\) khi \(x=y=\dfrac{1}{2}\)
\(1=2\left(\dfrac{9}{x}+\dfrac{1}{y}\right)\ge2.\dfrac{\left(3+1\right)^2}{x+y}=\dfrac{32}{x+y}\)
\(\Rightarrow x+y\ge32\)
\(A_{min}=32\) khi \(\left(x;y\right)=\left(24;8\right)\)
1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)
\(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)
max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)
\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t
Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)
Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có
\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)
\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)
\(A\ge\dfrac{\left(1+2\right)^2}{x+y}=9\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{3};\dfrac{2}{3}\right)\)
Tu de bai suy ra 2y+2x=xy<=>...<=>y(2-x)= -2x<=>y=2x/(x-2)<=>y=(2x-4+4)/(x-2)<=>y=2+4/(x-2)
vi x la so nguyen Dưỡng nen x-2 la so nguyen duong va la ước cua 4 => x-2 =1 hoặc x-2= 4 => x=3 hoac x=6
Voi x=3 => y= 6
voi x=6=> y=3
vay cac cap so nguyen duong (x;y) can tim la (3;6); (6;3)