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1, \(3xy^2+Q=-7xy^2\)
\(\Rightarrow Q=-7xy^2-3xy^2\)
\(\Rightarrow Q=-10xy^2\)
2. \(P=3+5x^2-3xy+5y-5x^2-11+2xy+x^3\)
\(\Rightarrow P=\left(5x^2-5x^2\right)+\left(-3xy+2xy\right)+5y+x^3+\left(3-11\right)\)
\(\Rightarrow P=-xy+5y+x^3-8\)
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Ta có: \(\hept{\begin{cases}\left(3x-2y\right)^{2020}\ge0;\forall x,y,z\\\left(5y-3z\right)^{2000}\ge0;\forall x,y,z\\|2z-5x|\ge0;\forall x,y,z\end{cases}}\)
\(\Rightarrow\left(3x-2y\right)^{2020}+\left(5y-3z\right)^{2000}+|2z-5x|\ge0;\forall x,y,z\)
Do đó \(\left(3x-2y\right)^{2020}+\left(5y-3z\right)^{2000}+|2z-5x|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(3x-2y\right)^{2020}=0\\\left(5y-3z\right)^{2000}=0\\|2z-5x|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x=2y\\5y=3z\\2z=5x\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\\\frac{z}{5}=\frac{x}{2}\end{cases}}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x+y-z}{2+3-5}=\frac{5}{0}\)( vô lý )
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