lm bừa == e thử sức thoi :))

\(\hept{\begin{cases}2x+\sqrt{3}y=\sqrt{3}\\\sqrt{2}x-3y=\sqrt{2}\end{cases}}\)

\(\hept{\begin{cases}-\sqrt{3}+\sqrt{3}y=2x\\\sqrt{2}x-\sqrt{2}=3y\end{cases}}\)

\(\hept{\begin{cases}-\sqrt{3}+\sqrt{3}y=2x\\3\sqrt{2}x-3\sqrt{2}=y\end{cases}}\)

\(\hept{\begin{cases}-\sqrt{3}+\sqrt{3}y=2x\\1=y\end{cases}}\)== y là bừa đó :(( 

\(\hept{\begin{cases}-\sqrt{3}+\sqrt{3}.1=2x\\y=1\end{cases}}\)

\(\hept{\begin{cases}0=2x\\y=1\end{cases}}\)

\(\hept{\begin{cases}x=0\\y=1\end{cases}}\)

Vậy pt có nghiệm là (x;y) = (0;1) 

Nhẩm lại thấy kết quả sai bn ak

Ta sẽ chứng minh:\(P\le\frac{5}{8}\Leftrightarrow5-8P=5+8abc-8\left(ab+bc+ca\right)\ge0\)

Ta có: \(5-8P=\frac{4ab\left[4\left(a+2bc-b-c\right)^2+\left(2c-1\right)^2\right]+c\left(2b-1\right)^2\left[4\left(a+b-c\right)^2+1\right]}{4ab+c\left(2b-1\right)^2}\ge0\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

Theo nguyên lý Dirichlet, trong ba số 2a - 1; 2b - 1; 2c - 1 tồn tại ít nhất hai số cùng dấu

Giả sử \(\left(2a-1\right)\left(2b-1\right)\ge0\Leftrightarrow4ab-2a-2b+1\ge0\)

\(\Leftrightarrow4abc\ge2ac+2bc-c\Leftrightarrow2abc\ge ac+bc-\frac{c}{2}\)

 Khi đó thì\(P=ab+bc+ca-2abc+abc\)\(\le ab+bc+ca-ac-bc+\frac{c}{2}+abc=ab+abc+\frac{c}{2}\)

\(\le\frac{a^2+b^2}{2}+abc+\frac{c}{2}=\frac{a^2+b^2+c^2+2abc}{2}-\frac{1}{2}\left(c^2-c+\frac{1}{4}\right)\)\(+\frac{1}{8}\)

\(=\frac{5}{8}-\frac{1}{2}\left(c-\frac{1}{2}\right)^2\le\frac{5}{8}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

\(\frac{x^2+2x+2}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow\frac{\left(x+2\right)^2-2}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow\left(x+2\right)^2-2=0\)Vì \(\left(x+1\right)^2\ne0\)

\(\Leftrightarrow\left(x+2\right)^2=2\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=\sqrt{2}\\x+2=-\sqrt{2}\end{cases}}\)

Để \(\frac{2x-4}{x+2}\)nguyên thì

\(2x-4⋮x+2\)

\(\Rightarrow2\left(x+2\right)-8⋮x+2\)

Mà \(2\left(x+2\right)⋮x+2\)

\(\Rightarrow8⋮x+2\)

\(\Rightarrow x+2\in\left\{1;2;4;8;-1;-2;-4;-8\right\}\)

\(\Rightarrow x\in\left\{-1;0;2;6;-3;-4;-6;-10\right\}\)

Học tốt

A=\(\frac{2x-4}{x+2}=\frac{2\left(x+2\right)-8}{x+2}=2-\frac{8}{x+2}\)

Để A nguyên thì \(\frac{8}{x+2}\)nguyên =>\(x+2\inƯ\left(8\right)=\left\{\pm1,\pm2,\pm4,\pm8\right\}\)

Ta có bảng:

Vậy x={-10,-6,-4,-3,1,0,2,6}thì A nguyên