mình làm cho bạn 2 cách nha

Cách 1 )

ta có \(1\le y\le2\Leftrightarrow\frac{1}{y^2+1}\ge\frac{1}{2x+3}\)

ta có \(xy+2\ge2y\Leftrightarrow x\ge\frac{2\left(y-1\right)}{y}\ge0\)

ta có \(M=\frac{x^2+4}{y^2+1}=\left(x^2+4\right).\frac{1}{y^2+1}\ge\left(2x+3\right).\frac{1}{2x+3}=1\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

zậy \(minM=\frac{x^2+4}{y^2+1}khi\hept{\begin{cases}x=1\\y=2\end{cases}}\)

cách 2)

ta có \(1\le y\le2;xy+2\ge2y\Leftrightarrow4xy+8\ge8y;4x^2+y^2+8\ge4xy+8\)

từ đó ta có

\(4\left(x^2+4\right)\ge-y^2+8+8y=4\left(y^2+1\right)+\left(5y+2\right)\left(2-y\right)\ge4\left(x^2+1\right)\Rightarrow M=1\)

zậy kết luận như cách 1

\(\left(x^2-x+1\right)\left(xy+y^2\right)=3x-1\left(1\right)\)

\(3x-1⋮x^2-x+1\)

zì \(lim\left(x\rightarrow\infty\right)\frac{3x-1}{x^2-x+1}=0\)

zà thấy x=2 thỏa mãn ,=> x=1

thay zô 1 ta có

\(1\left(y+y^2\right)=2=>y^2+y-2=0=>\orbr{\begin{cases}y=1\\y=-2\end{cases}}\)

zậy \(\left(x,y\right)\in\left\{\left(1,1\right)\left(1,-2\right)\right\}\)

Với m =1 suy ra : 

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

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

\(\Rightarrow\hept{\begin{cases}y=2.3-1=5\\x=3\end{cases}}\)

b ) Để hệ có nghiệm x+2y=3 

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

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

\(\Rightarrow\hept{\begin{cases}x=3-2.\frac{5}{3}=-\frac{1}{3}\\y=\frac{5}{3}\end{cases}}\)

\(\Rightarrow2.\left(-\frac{1}{3}\right)-\frac{5}{3}=2m-1\Rightarrow m=-\frac{2}{3}\)

con chó\

Ta có : 

\(P=a+2b=\left(a-2\right)+2\left(b+3\right)-4\)

\(\Rightarrow P+4=\left(a-2\right)+2\left(b+3\right)\)

\(\Rightarrow\left(P+4\right)^2=\left(\left(a-2\right)+2\left(b+3\right)\right)^2\le\left(1^2+2^2\right)\left(\left(a-2\right)^2+\left(b+3\right)^2\right)\)

\(=25\)

\(\Rightarrow-5\le P+4\le5\)

\(\Rightarrow P\ge-9\)

Dấu " = " xảy ra khi \(\frac{a-2}{1}=\frac{b+3}{2},\left(a-2\right)^2+\left(b+3\right)^2=5\)

\(\Rightarrow a-2=-1,b+3=-2\Rightarrow a=1,b=-5\)

Chứng minh \(\frac{1}{\left(x+y\right)\left(x+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)

\(\frac{1}{2x+y+z}=\frac{1}{\left(x+y\right)\left(x+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

                                                                \(\le\frac{1}{4}.\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)\)

                                                                  \(\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

 tương tự cộng zế zới zế ta đc

\(\frac{1}{2xx+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)

                                                                                      \(\le\frac{1}{16}.4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)(dpcm