x²-2+\(\frac{1}{x^2}\) +x²-xy+\(\frac{y^2}{4}=2-xy\)

=>\(\left(x-\frac{1}{x}\right)^2+\left(x-\frac{y}{2}\right)^2=2-xy\)

Do VT\(\ge0\)=> 2-xy\(\ge0\)

                       =>xy\(\le2\)

 Vậy Max_xy=2 (dấu bằng tự làm)

à mình đọc nhầm tưởng là gtln.

 \(x^2-2+\frac{1}{x^2}+x^2\)\(+xy+\frac{y^2}{4}=2+xy\)

=>\(\left(x-\frac{1}{x}\right)^2+\left(x+\frac{y}{2}\right)^2\)=2+xy

Do VT\(\ge0\)=> 2+xy\(\ge0\)

                      =>xy\(\ge-2\)

Vậy Min_xy=2

Min = -2 nhé ghi nhầm

x^2 -2 +1/x^2+x^2+xy+y^2/4=2+xy

(x-1/x)^2+(x+y/2)^2=2+xy

suy ra được min xy=-2 khi x=1,y=-2

Cao nhân nào giải được bài này chưa

Áp dụng bđt bunhia cho 2 bộ số \(\left(\frac{a}{x};\frac{b}{y}\right),\left(x;y\right)\)ta được

\(\left(\frac{a}{x}+\frac{b}{y}\right)\left(x+y\right)\ge\left(\sqrt{\frac{a}{x}.x}+\sqrt{\frac{b}{y}.y}\right)^2\)

\(\rightarrow x+y\ge\left(\sqrt{a}+\sqrt{b}\right)^2\)

\(MinS=\left(\sqrt{a}+\sqrt{b}\right)^{22}\)

1. Ta có \(1+x^2\ge2x\), \(1+y^2\ge2y\), \(1+z^2\ge2z\)

Suy ra \(P=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)

Chọn D. \(P\le\frac{1}{2}\)

2. a) Áp dụng BĐT Bunhiacopxki, ta có

\(\left(\frac{1}{x}+\frac{4}{y}\right)\left(x+y\right)\ge\left[\left(\sqrt{\frac{1}{x}.x}\right)^2+\left(\sqrt{\frac{4}{y}.y}\right)^2\right]=\left(1^2+2^2\right)\)

\(\Rightarrow\frac{1}{x}+\frac{4}{y}\ge1\)

Đẳng thức xảy ra khi \(\left\{\begin{matrix}\frac{1}{x^2}=\frac{4}{y^2}\\x+y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}x=\frac{10}{3}\\y=\frac{5}{3}\end{matrix}\right.\)

2.b

\(\left|x\right|.\sqrt{1-x^2}=\sqrt{x^2\left(1-x^2\right)}\le\frac{x^2+1-x^2}{2}\)

\(\Rightarrow\left|x\right|\sqrt{1-x^2}\le\frac{1}{2}\)

hay \(-\frac{1}{2}\le x\sqrt{1-x^2}\le\frac{1}{2}\)

Bạn tự tìm được rồi nhé :)

\(C=\frac{30}{4x-4x^2-6}=\frac{-30}{4x^2-4x+6}=\frac{-30}{\left(2x-1\right)^2+5}\)

Vì \(\left(2x-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2+5\ge5\Rightarrow\frac{1}{\left(2x-1\right)^2+5}\le\frac{1}{5}\Rightarrow C=\frac{-30}{\left(2x-1\right)^2+5}\ge\frac{-30}{5}=-6\)

Dấu "=" xảy ra khi x=1/2

Vậy Cmin=-6 khi x=1/2

\(E=\frac{1000}{x^2+y^2-20x-20y+2210}=\frac{1000}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\)

Vì \(\left(x-10\right)^2\ge0;\left(y-10\right)^2\ge0\Rightarrow\left(x-10\right)^2+\left(y-10\right)^2\ge0\)

\(\Rightarrow\left(x-10\right)^2+\left(y-10\right)^2+2010\ge2010\)

\(\Rightarrow\frac{1}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\le\frac{1}{2010}\)

\(\Rightarrow E=\frac{1000}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\le\frac{1000}{2010}=\frac{100}{201}\)

Dấu "=" xảy ra khi x=y=10

Vậy Emax = 100/201 khi x=y=10

Ta có: \(E=\frac{1000}{x^2+y^2-20x-20y+2210}=\frac{1000}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\)

Vì\(\left(x-10\right)^2\ge0;\left(y-10\right)^2\ge0\Rightarrow\left(x-10\right)^2+\left(y-10\right)^2\ge0\)

\(\Rightarrow\left(x-10\right)^2+\left(y-10\right)^2+2010\ge2010\)

\(\Rightarrow\frac{1}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\le\frac{1}{2010}\)

\(\Rightarrow E=\frac{1000}{\left(x-10\right)^2+\left(y-10\right)^2+2010}\le\frac{1000}{2010}=\frac{100}{201}\)

Dấu "=" xảy ra khi \(x=y=10\)

Vậy Emax\(=\frac{100}{201}\)khi \(x=y=10\)