Ta có: \(\left(x-1\right)^{2020}\ge0\forall x\)

\(\left|y-3\right|\ge0\forall y\)

Do đó: \(\left(x-1\right)^{2020}+\left|y-3\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

Vậy: (x,y)=(1;3)

Sửa đề: \(E=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)

-x+y-z=0

=>x-y+z=0

=>x+z=y; x-y=-z; -y+z=-x

=>x+z=y; x-y=-z; y-z=x

Ta có: \(E=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)

\(=\frac{x+z}{z}\cdot\frac{x-y}{x}\cdot\frac{y-z}{y}\)

\(=\frac{y}{z}\cdot\frac{-z}{x}\cdot\frac{x}{y}=-1\)

\(\Leftrightarrow y\left(x+1\right)+2\left(x+1\right)+9=0\)

\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=-9\)

Để x;y nguyên thì:

\(\left\{{}\begin{matrix}x+1=3\\y+2=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=-3\\y+2=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=1\\y+2=-9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-11\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=-9\\y+2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=-1\\y+2=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=9\\y+2=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)

mình ko bt nx thầy mình tự nghĩ cho bọn mình làm thôi

Hình như x=6 y=7 vì 6/6=1 7/7=1 nên x/6 = y/7

nhưng y lại lớn hơn 0 mà

bạn ơi có câu c không bạn

\(x^2-4x+3=0\\ \Rightarrow\left(x^2-3x\right)-\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)-\left(x-3\right)=0\\ \Rightarrow\left(x-1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

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