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\(B=\frac{ab}{a+b+2}\Rightarrow2B=\frac{2ab}{a+b+2}=\frac{\left(a+b\right)^2-a^2-b^2}{a+b+2}=\frac{\left(a+b\right)^2-4}{a+b+2}=a+b-2\)
Do a ; b không âm , áp dụng BĐT Cô - si cho 2 số , ta có :
\(a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2.4}=\sqrt{8}\)
\(\Rightarrow a+b-2\le\sqrt{8}-2\)
\(\Rightarrow2B\le\sqrt{8}-2\Rightarrow B\le\sqrt{2}-1\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=\sqrt{2}\)
Do x ; y không âm , \(x^2+y^2=1\)
\(\Rightarrow\left|x\right|;\left|y\right|\le1\) \(\Rightarrow0\le x;y\le1\)
\(\Rightarrow x\ge x^2;y\ge y^2\Rightarrow x+y\ge x^2+y^2=1\)
\(x,y\ge0\Rightarrow xy\ge0\)
Ta có : \(A=\sqrt{5x+4}+\sqrt{5y+4}\)
\(\Rightarrow A^2=5x+4+5y+4+2\sqrt{\left(5x+4\right)\left(5y+4\right)}\)
\(=5\left(x+y\right)+8+2\sqrt{25xy+20y+20x+16}\)
\(\ge5.1+8+2\sqrt{25.0+20.1+16}=13+2.6=25\)
\(\Rightarrow A\ge5\)
Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)
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
\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\)
\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)
Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)
\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)
minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)
maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)
\(A=x-2y+3z\left(x,y,z>0\right)\)
\(\left\{{}\begin{matrix}2x+4x+3z=8\left(1\right)\\3x+y-3z=2\left(2\right)\end{matrix}\right.\)
(1) <=> \(5x+5y=10\) <=> x+ y = 2
=> y = 2-x
Từ (1) => \(2x+4\left(2-x\right)+3z=8\)
=> -2x +3z =0
=> \(x=\dfrac{3}{2}z\) => \(z=\dfrac{2}{3}x\) thay vào A
=> \(A=x-2\left(2-x\right)+3.\dfrac{2}{3}x=5x-4\ge-4\)
Vậy Amin = -4.
Ta có :
x*x+y*y=2