\(\Leftrightarrow\left\{{}\begin{matrix}3x\left(2x-y\right)+x+y-1=0\\x^2+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\left(2x-y\right)-\left(2x-y\right)+3x-1=0\\x^2+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-1\right)\left(2x-y\right)+3x-1=0\\x^2+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-1\right)\left(2x-y+1\right)=0\\x^2+y^2=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải hpt \(\left\{{}\begin{matrix}\frac{x^2}{y^2+2y+1}+\frac{y^2}{x^2+2x+1}=\frac{1}{2}\\3xy-x-y=1\end{matrix}\right.\)
Giai hệ phương trình:
1,\(\left\{{}\begin{matrix}\sqrt{x+1}+\sqrt{y}=4\\x+y=7\end{matrix}\right.\) 2,\(\left\{{}\begin{matrix}x^2+y^{^2}-xy=1\\x^3+y^3=x+3y\end{matrix}\right.\) 3,\(\left\{{}\begin{matrix}6x^2-3xy+x+y=1\\x^2+y^2=1\end{matrix}\right.\)
10. giải hpt bằng phương pháp thế:
6) \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}3x-2=y\\2x+3y=6\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}2x+3y=2\\4x-y-1=0\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-2y=3\\2x-\dfrac{4}{3}y=1\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}5x+y=3\\2x+0,4y=1,2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
Giải hpt
\(\left\{{}\begin{matrix}x^3-2y^3+3\left(x-2y\right)=3xy\left(x-y\right)\\2x^3=\left(1+4y-3x^2\right)\sqrt{2x+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6x^2-3xy+x=1-y\\x^2+y^2=1\end{matrix}\right.\)
Giải hpt \(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
Giúp mk với! Cảm ơn ạ!
Giải các hệ PT sau:
a) \(\left\{{}\begin{matrix}2x^2-3xy=y^2-3x-1\\2y^2-3xy=x^2-3y-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^3-2y=4\\y^3-2x=4\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\sqrt{x+1}-\sqrt{7-y}=4\\\sqrt{y+1}-\sqrt{7-x}=4\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}2x^2=y+\frac{1}{y}\\2y^2=x+\frac{1}{x}\end{matrix}\right.\)
giải hpt \(\left\{{}\begin{matrix}2\left(x-2\right)^2+\dfrac{1}{\sqrt{y+5}}=3\\\left(x-2\right)^2-\dfrac{2}{\sqrt{y+5}}=-1\end{matrix}\right.\)
giải hpt:
\(\left\{{}\begin{matrix}6x+6y=5xy\\\dfrac{4}{x}+\dfrac{3}{y}=1\end{matrix}\right.\)
