Mình sửa lại đề cho đúng nhé

\(\hept{\begin{cases}3x-y=3z\\2x+y=7z\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2z\\y=3z\end{cases}}\)

Thế vô M ta được

\(M=\frac{x^2-2xy}{x^2+y^2}=\frac{4z^2-2.2z.3z}{4z^2+9z^2}=-\frac{8}{13}\)

\(_{\hept{\begin{cases}3x-y=3z\left(1\right)\\\\2x+y=7z\end{cases}\Rightarrow}\left(3x-y\right)+\left(2x+y\right)=10z}\)

\(\Leftrightarrow\)5x=10z\(\Leftrightarrow x=2z\)

thay x=2z vào (1) ta được :6z+y=3z\(\Rightarrow y=6z-3z=3z\)

thay y=3z,x=2z vào biểu thức M=\(\frac{4z^2-12z^2}{4z^2+9z^2}=\frac{-8}{13}\)

ta có 5x=10z=> x=2z=> y=3z

Tháy vào, ta có \(M=\frac{4z^2-12z^2}{4z^2+9z^2}=\frac{-8z^2}{13z^2}=-\frac{8}{13}\)

Ta có:

\(3x-y+2x+y=3z+7z\) 

\(5x=10z\) 

\(x=2z\) 

thay:\(4z+y=7z\) \(\Rightarrow y=3z\) 

Thay vào M ta đc:M=\(\frac{4z^2-12z^2}{4z^2+9z^2}\) =\(\frac{-8z^2}{13z^2}=\frac{-8}{13}\) 

vậy\(M=\frac{-8}{13}\) nếu\(3x-y=3z;2x+y=7z\) 

\(3x-y=3z\Rightarrow-y=3z-3x\Rightarrow y=3x-3z\)

\(2x+y=7z\Rightarrow y=7z-2x\)\(\Rightarrow3x-3z=7z-2x=y\Rightarrow3x-3z-7z+2x=5x-10z=0\Rightarrow x-2z=0\Rightarrow x=2z\)

\(2x+y=7z\Rightarrow2\cdot2z+y=7z\Rightarrow4z+y=7z\Rightarrow y=3z\)

\(M=\frac{x^2-2xy}{x^2+y^2}=\frac{\left(2z\right)^2-2\cdot2z\cdot3z}{\left(2z\right)^2+\left(3z\right)^2}=\frac{4z^2-12z^2}{4z^2+9z^2}=-\frac{8z^2}{13z^2}=-\frac{8}{13}\)

Ta có: \(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Leftrightarrow3x-y+2x+y=10z\)

\(\Leftrightarrow5x=10z\)

hay x=2z

Thay x=2z vào biểu thức 3x-y=3z, ta được:

\(3\cdot2z-y=3z\)

\(\Leftrightarrow6z-y=3z\)

hay y=3z

Thay x=2z và y=3z vào biểu thức \(M=\dfrac{x^2-2xy}{x^2+y^2}\), ta được:

\(M=\dfrac{\left(2z\right)^2-2\cdot2z\cdot3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{13z^2}=\dfrac{-8z^2}{13z^2}=\dfrac{-8}{13}\)

Vậy: \(M=\dfrac{-8}{13}\)

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5x=10z\\3x-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\3.2z-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=3.2z-3z=6z-3z=3z\end{matrix}\right.\)

Có: \(M=\dfrac{x^2-2xy}{x^2+y^2}=\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{4z^2+9z^2}=\dfrac{-8z^2}{13z^2}==-\dfrac{8}{13}\)

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5x=10z\\3x-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\3.2z-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=3.2z-3z=6z-3z=3z\end{matrix}\right.\)

Có: \(M=\dfrac{x^2-2xy}{x^2+y^2}=\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{4z^2+9z^2}=\dfrac{-8z^2}{13z^2}==-\dfrac{8}{13}\)

Ta có:

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}5x=10z\\2x+y=7z\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x=2z\\y=3z\end{matrix}\right.\)

Thay x = 2z và y = 3z vào biểu thức M ta được:

M = \(\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}\)

= \(\dfrac{4z^2-12z^2}{4z^2+9z^2}\)

= \(\dfrac{-8z^2}{13z^2}\)

= \(\dfrac{-8}{13}\)

Vậy...

a)

\(x^3+y^3+3\left(x^2+y^2\right)+4\left(x+y\right)+4=0\)

\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)+\left(y^3+3y^2+3y+1\right)+\left(x+y+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+\left(x+y+2\right)=0\)

\(\Leftrightarrow\left(x+y+2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\right]+\left(x+y+2\right)=0\)

\(\Leftrightarrow\left(x+y+2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2+1\right]=0\)

Lại có :\(\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2+1=\left[\left(x+1\right)-\frac{1}{2}\left(y+1\right)\right]^2+\frac{3}{4}\left(y+1\right)^2+1>0\)

Nên \(x+y+2=0\Rightarrow x+y=-2\)

Ta có :

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

Vì \(4xy\le\left(x+y\right)^2\Rightarrow4xy\le\left(-2\right)^2\Rightarrow4xy\le4\Rightarrow xy\le1\)

\(\Rightarrow\frac{1}{xy}\ge\frac{1}{1}\Rightarrow\frac{-2}{xy}\le-2\)

hay \(M\le-2\)

Dấu "=" xảy ra khi \(x=y=-1\)

                    Vậy \(Max_M=-2\)khi \(x=y=-1\)

c)  ( Mình nghĩ bài này cho x, y, z ko âm thì mới xảy ra dấu "=" để tìm Min chứ cho x ,y ,z dương thì ko biết nữa ^_^  , mình làm bài này với điều kiện x ,y ,z ko âm nhé )

Ta có :

\(\hept{\begin{cases}2x+y+3z=6\\3x+4y-3z=4\end{cases}\Rightarrow2x+y+3z+3x+4y-3z=6+4}\)

\(\Rightarrow5x+5y=10\Rightarrow x+y=2\)

\(\Rightarrow y=2-x\)

Vì \(y=2-x\)nên \(2x+y+3z=6\Leftrightarrow2x+2-x+3z=6\)

\(\Leftrightarrow x+3z=4\Leftrightarrow3z=4-x\)

\(\Leftrightarrow z=\frac{4-x}{3}\)

Thay \(y=2-x\)và \(z=\frac{4-x}{3}\)vào \(P\)ta có :

\(P=2x+3y-4z=2x+3\left(2-x\right)-4.\frac{4-x}{3}\)

\(\Rightarrow P=2x+6-3x-\frac{16}{3}+\frac{4x}{3}\)

\(\Rightarrow P=\frac{x}{3}+\frac{2}{3}\ge\frac{2}{3}\)( Vì \(x\ge0\))

Dấu "=" xảy ra khi \(x=0\Rightarrow\hept{\begin{cases}y=2\\z=\frac{4}{3}\end{cases}}\)( Thỏa mãn điều kiện y , z ko âm )

Vậy \(Min_P=\frac{2}{3}\)khi \(\hept{\begin{cases}x=0\\y=2\\z=\frac{4}{3}\end{cases}}\)

mik cx ko chắc vs câu c

ib tui làm cho