3(x - 3) = - 2(2 - x)

3x - 9 = -4 - ( -2x )

3x - 9 = -4 + 2x

3x - 2x = -4 +9

x = 5

\(x+y=1\Rightarrow2\sqrt{xy}\le1\Rightarrow\sqrt{xy}\le\frac{1}{2}\)

\(\Rightarrow xy\le\frac{1}{4}\Rightarrow\frac{1}{xy}\ge4\)

Áp dụng bđt cauchy cho 3 số dương:

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{xy}\ge3\sqrt[3]{\frac{1}{x^2}.\frac{1}{y^2}.\frac{1}{xy}}=3.\frac{1}{xy}\ge3.4=12\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)

Làm tiếp ạ

\(\Rightarrow P\ge\frac{289}{16}\)

Dấu"="Xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

Vậy MIN P=\(\frac{289}{16}\)\(\Leftrightarrow x=y=\frac{1}{2}\)

Em chả có cách gì ngoài cô si mù mịt :v

\(\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)\)

\(=\left(x^2+\frac{1}{16y^2}+\frac{1}{16y^2}+.....+\frac{1}{16y^2}\right)\left(y^2+\frac{1}{16x^2}+\frac{1}{16x^2}+.....+\frac{1}{16x^2}\right)\)

\(\ge17\sqrt[17]{\frac{x^2}{16^{16}\cdot y^{32}}}\cdot17\sqrt[17]{\frac{y^2}{16^{16}\cdot x^{32}}}\)

\(=17^2\sqrt[17]{\frac{x^2y^2}{16^{32}\cdot x^{32}\cdot y^{32}}}\)

\(=17^2\sqrt[17]{\frac{1}{16^{32}\cdot\left(xy\right)^{30}}}\)

\(\ge17^2\sqrt[17]{\frac{1}{16^{32}\left(\frac{x+y}{2}\right)^{60}}}=\frac{289}{16}\)

Dấu "=" xảy ra tại x=y=¹/₂

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

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

\(=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

Áp dụng BĐT AM-GM ta có:

\(x^2y^2+\frac{1}{256x^2y^2}\ge2\sqrt{x^2y^2.\frac{1}{256x^2y^2}}=\frac{1}{8}\)

\(xy\le\frac{\left(x+y\right)^2}{4}\Rightarrow x^2y^2\le\frac{\left(x+y\right)^4}{16}=\frac{1}{16}\)

\(\Rightarrow P\ge\frac{1}{8}+\frac{255}{256.\frac{1}{16}}+2\)

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

\(\Leftrightarrow xy\ge4\)

\(\Rightarrow A=xy+2017\ge4+2017=2021\)

Ta có: \(\frac{xy+1}{x+y}\ge\frac{3y+1}{x+y}\ge\frac{3y+1}{2y}>\frac{3y}{2y}=\frac{3}{2}\)( mâu thuẫn với gt)

giả sử \(a\le2\Rightarrow a\in\left\{1;2\right\}\)

+ Với a=1 \(\Rightarrow M=\frac{y^3+1}{y^3+1}=1\)

+ Với a=2 \(\Rightarrow M=\frac{8y^3+1}{y^3+8}\)

Từ đk \(\frac{xy+1}{x+y}=\frac{2y+1}{y+2}< \frac{3}{2}\Rightarrow b< 4\)

=> \(b\in\left\{1;2;3\right\}\)

+ Với b=1 \(\Rightarrow M=\frac{9}{9}=1\)

+ Với  b=2 \(\Rightarrow M=\frac{8.8+1}{8+8}=\frac{65}{16}\)

+ vỚI b=3 \(\Rightarrow M=\frac{8.27+1}{27+8}=\frac{217}{35}\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\) hoặc ngược lại.

Ta có: \(\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)

=> \(\left|2x+3\right|+\left|2x-1\right|\ge4\)(1)

Ta lại có: \(\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)

=> \(\left|2x+3\right|+\left|2x-1\right|\ge4\) (2)

Từ (1); (2) : \(\left|2x+3\right|+\left|2x-1\right|=\frac{8}{3\left(x+1\right)^2+2}\)

<=> \(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\\\left(x+1\right)^2=0\end{cases}\Leftrightarrow x=-1}\)(TM)

Vậy:... 

\(-\frac{\left(-x\right)}{5}-\frac{2}{10}=\frac{1}{-5}-\frac{7}{50}\)

\(\Rightarrow\frac{x}{5}-\frac{2}{10}=-\frac{17}{50}\)

\(\Rightarrow\frac{x}{5}=-\frac{17}{50}+\frac{2}{10}\)

\(\Rightarrow\frac{x}{5}=-\frac{7}{50}\)

\(\Rightarrow x\in\theta\)

tíc mình nha