\(\left\{{}\begin{matrix}x-y=2\left(1\right)\\x^2+y^2=10\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=y+2\)
Thay vào \(\left(2\right)\Leftrightarrow\left(y+2\right)^2+y^2=10\Leftrightarrow y^2+2y-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-3\Rightarrow x=-1\\y=1\Rightarrow x=3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;-3\right);\left(3;1\right)\)
giải hệ:
\(\left\{{}\begin{matrix}x+2y=7\\x^2+y^2-2xy=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y=2\\x^2+y^2+164\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y+xy=-13\\x^2+y^2-x-y=32\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y=3\\x^3-y^3=7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+2y=10\\\dfrac{1}{2}x-y=1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x^2-y^2=3\left(x-y\right)\\xy=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\sqrt{y}+y\sqrt{x}=6\\x^2y+y^2x=20\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=10\\\left(x+1\right)\left(y+1\right)=8\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y+3xy=-3\\xy+1=0\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}x^2-y^2=16\\x+y=8\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=10\\x+y=4\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}8y-x=4\\2x-21y=2\end{matrix}\right.\) b)\(\left\{{}\begin{matrix}x+y=-2.\left(x-1\right)\\7x+3y=x+y+5\end{matrix}\right.\)
giải hệ pt
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\left(x-1\right)+\left(y+2\right)=2\\4\left(x-1\right)+3\left(y+2\right)=7\end{matrix}\right.\)
