tìm nghiệm nguyên
\(5x+25=-3xy+8y^2\)
Giải hpt : a) \(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+1\right)=25\left(y+1\right)\\x^2+xy+2y^2+x-8y=9\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}x^2+y^2+6xy-\frac{1}{\left(x-y\right)^2}+\frac{9}{8}=0\\2y-\frac{1}{x-y}+\frac{5}{4}=0\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{x}{x^2-y}+\frac{5y}{x+y^2}=4\\5x+y+\frac{x^2-5y^2}{xy}=5\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}3xy+y+1=21x\\9x^2y^2+3xy+1=117x^2\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=1\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+3xy-3\left(x-y\right)=0\\x^4+9y\left(x^2+y\right)-5x^2=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{3y+1}+\sqrt{5x+4}=3xy-y+3\\\sqrt{2x^{^2}+2y^{^2}}+\sqrt{\frac{4}{3}\left(x^{^2}+y^{^2}+xy\right)}=2\left(x+y\right)\end{matrix}\right.\)
5x+25=−3xy+8y2
Giải hệ pt:
\(\left\{{}\begin{matrix}x^3+y^3+3xy=1\\\sqrt{\left(4-x\right)\left(13-y\right)}=\dfrac{2x+2y+25}{2x+y+2}\end{matrix}\right.\)
Giải hệ phương trình : \(\left\{{}\begin{matrix}x^2+y^2+x+y=8\\2x^2+y^2-3xy+3x-2y+1=0\end{matrix}\right.\)
Giải hệ pt: (Em xin lỗi vì bài trước em ghi sai đề:()
\(\left\{{}\begin{matrix}x^3+y^3+3xy=1\\\sqrt{\left(4-x\right)\left(13-y\right)}=\dfrac{2x+3y+25}{2x+y+2}\end{matrix}\right.\)
Giải hệ : \(\left\{{}\begin{matrix}x+y+z=8\\3xy+z^2=45\end{matrix}\right.\)