giải hệ phương trình :
\(\left\{{}\begin{matrix}x^2+y^2-2x+2y-2=0\\x^2+\left(3y+4\right)y=2\left(2-x-y\right)\left(1+xy\right)\end{matrix}\right.\)
giải hệ phương trình :
\(\left\{{}\begin{matrix}x^2+y^2-2x+2y-2=0\\x^2+\left(3y+4\right)y=2\left(2-x-y\right)\left(1+xy\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy+x-y=3\\x^2+y^2+x+y=12\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}2x^2+y^2-3xy-4x+3y+2=0\\\sqrt{x^2-y+3}+\sqrt{y-x+1}=2\end{matrix}\right.\)
\(\sqrt{x^2-y+3}+\sqrt{y-x+1}=2\)
Xét \(pt\left(1\right)\Leftrightarrow2x^2+y^2-3xy-4x+3y+2=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(2x-y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=2x-2\end{matrix}\right.\)
*)\(y=x-1\) thay vao \(pt(2)\) :
\(pt\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=0\end{matrix}\right.\)
*)\(y=2x-2\) thay vao \(pt(2)\):
\(pt\Leftrightarrow\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{1}{\sqrt{x-1}}\right)=0\)
\(\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)
\(\left\{{}\begin{matrix}\left(x^2+x+1\right)\left(y^2+y+1\right)=3\\\left(1-x\right)\left(1-y\right)=6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=1\\x^3+y^3=x^2+y^2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
bạn nào biết làm câu nào thì giúp mik nha☺️☺️☺️ thanks
1) \(\left\{{}\begin{matrix}x+y=1\\x^3+y^3=x^2+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x^2+y^2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x^2-xy+y^2-x^2-y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=1\\y=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\\left(x^2+y^2\right)^2-3x^2y^2=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\\left(5\right)^2-3x^2y^2=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\-3x^2y^2=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\x^2y^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\\left(5-y^2\right)y^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\-y^4+5y^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\-y^4+5y^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\\left[{}\begin{matrix}y^2=1\\y^2=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=5-y^2\\y^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=5-y^2\\y^2=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=4\\y^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=1\\y^2=4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
vậy \(S=\left\{\left(2;1\right),\left(2;-1\right),\left(-2;1\right),\left(-2;-1\right),\left(1;2\right),\left(1;-2\right),\left(-1;2\right),\left(-1;-2\right)\right\}\)
4) \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5-1=-y^5\\x^9-x^4+y^9-y^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x^5-1=-y^5\\y^5-1=-x^5\end{matrix}\right.\\x^4\left(x^5-1\right)+y^4\left(y^5-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\x^4\left(-y^5\right)+y^4\left(-x^5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\-x^4y^4\left(x+y\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\\left[{}\begin{matrix}x=0\\y=0\\x+y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^5+y^5=1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^5+y^5=1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^5+y^5=1\\x=-y\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-y\\x^5-x^5=1\end{matrix}\right.\end{matrix}\right.\)
vậy \(S=\left\{\left(1;0\right),\left(0;1\right)\right\}\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(1+\dfrac{1}{xy}\right)=4\\xy+\dfrac{1}{xy}+\dfrac{x^2+y^2}{xy}=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2+x+y=4\\x\left(x+y+1\right)+y\left(y+1\right)=2\end{matrix}\right.\)
từ pt (2) -> xy=2-x(x+1)-y(y+1)=2-4=-2.
từ pt (1) ->x2+y2-4+x+y=0. Thay xy=-2 vào ta được:
x2+y2+2xy+x+y=0<=>x+y=0 hoặc x+y=-1
Nếu x+y=0->x=-y ->y2=2 ->y=\(\pm\sqrt{2}\)->x=\(\mp\sqrt{2}\)
Nếu x+y=-1->\(\left[{}\begin{matrix}x=1;y=-2\\x=-2;y=1\end{matrix}\right.\)
Vậy (x;y)\(\in\left\{\left(\sqrt{2};-\sqrt{2}\right);\left(-\sqrt{2};\sqrt{2}\right);\left(1;-2\right);\left(2;-1\right)\right\}\)
\(\left\{{}\begin{matrix}xy^2-x^2y-y^3+5x^2+y^2-y+1=0\\x^2-y-3x+2=0\end{matrix}\right.\)
\(-x+3=2\sqrt{1-x}-\sqrt{1+x}+3\sqrt{1-x^2}\)
giải hệ :
\(\left\{{}\begin{matrix}2+6y=\dfrac{x}{y}-\sqrt{x-2y}\\\sqrt{x+\sqrt{x-2y}}=x+3y-2\end{matrix}\right.\)
Lời giải: https://hoc24.vn/hoi-dap/question/905017.html
Giải giúp mình hệ phương trình này với:
\(\left\{{}\begin{matrix}\dfrac{y-2x\: +\: \sqrt{y}\: -\: x}{\sqrt{xy}}+ 1=\: 0\\\sqrt{1\: -\: xy}\: +x^2\: -\: y^2\: =\: 0\end{matrix}\right.\)