Tọa độ giao điểm của (d1) và (d2) là:
\(\left\{{}\begin{matrix}2x+3=x+5\\y=x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=7\end{matrix}\right.\)
Thay x=2 và y=7 vào (d3), ta được:
4k-5=7
hay k=3
giải hệ
a,\(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=15\\\left(x-y\right)\left(x^2-y^2\right)=3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^3-y^3=9\\x^2+2y^2=x-4y\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\left(x-y\right)\left(2x+3y\right)=12\\6\left(x-y\right)+xy\left(x-y\right)=12\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+y^2+1=2\left(x+y\right)\\y\left(2x-y\right)=\left(2y+1\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{1}{x}-\frac{1}{y-2}=-1\\\frac{4}{x}+\frac{3}{y-2}=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{1}{x}+2\left(x+y\right)=3\\3x\left(x+y\right)-x=2\end{matrix}\right.\)
giải hệ:
a) \(\left\{{}\begin{matrix}\sqrt{x+3y}+\sqrt{x+y}=2\\\sqrt{x+y}+y-x=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left(x-\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=2\\2x^2y+xy^2-4xy=2x-y\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}2x^2+xy=y^2-3y+2\\x^2-y^2=3\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}x^2+y^2+z^2+2xy-xz-zy=3\\x^2+y^2-2xy-xz+zy=-1\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}x^2-y^2+5x-y+6=0\\x^2+\left(x-y\right)^2=2+\sqrt{6x+7}+2\sqrt{x+y+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\left(x+1\right)-5\left(y+1\right)=8\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\sqrt{5}-\left(1+\sqrt{3}\right)y=1\\\left(1+\sqrt{3}\right)x+y\sqrt{5}=1\end{matrix}\right.\)giải hệ phương trình
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
giải hệ phương trình sau
\(\left\{{}\begin{matrix}x^3+x^2\left(y-1\right)-5\left(x+y\right)=5\\3\sqrt{1+2x^2}+2\sqrt{40+9y^2}=5\sqrt{11}\end{matrix}\right.\)
Giải hệ phương trình:
a) \(\left\{{}\begin{matrix}\left|x+1\right|+\sqrt{y}=5\\\left(x^2+2x+1\right)y=36\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\2\left(x+y\right)-3\sqrt{x+1}=-5\end{matrix}\right.\)
