\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2xy=-2\\\left(x+y\right)^3-3xy\left(x+y\right)=8\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+2b=-2\\a^3-3ab=8\end{matrix}\right.\)
\(\Rightarrow a^3-3a\left(\frac{-2-a}{2}\right)=8\)
\(\Leftrightarrow2a^3+3a^2+6a-16=0\)
Rất tiếc rằng pt này ko giải được nên có vẻ đề bài ko đúng
Nếu đổi pt đầu thành \(x+y+2xy=2\) thì hoàn toàn giải được, có vẻ bạn chép nhầm dấu
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^2+2xy^2=3\\y^3+y+x\left(2xy-1\right)=3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+x^3y-xy^2+xy-y=1\\x^4+y^2-xy\left(2x-1\right)=1\end{matrix}\right.\)
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^2+8y^2=12\\x^3+2xy^2+12y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^3+y^3=1\\x^7+y^7=\left(x^4+y^4\right).1\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?
Giúp mình với, thanks các bạn nhiều: ^^ BT/ Giải hệ pt:
1/\(\left\{{}\begin{matrix}x^3+y^3=1\\x^2y+2xy^2+y^3=2\end{matrix}\right.\) 2/\(\left\{{}\begin{matrix}y^2=\left(x+8\right).\left(x^2+2\right)\\y^2-4\left(x+2\right)y+16+16x-5x^2=0\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}x^2-3x\left(y-1\right)+y^2+y\left(x-3\right)=4\\x-xy-2y=1\end{matrix}\right.\) 3/\(\left\{{}\begin{matrix}\sqrt{x}-\sqrt{x-y-1}=1\\y^2+x+2y\sqrt{x}-xy^2=0\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x^2+\left(3y+1\right)x+2y^2+y=0\\x^2+y^2+x+y=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}xy-x+y=3\\x^2+y^2-x+y+3xy=12\end{matrix}\right.\)
help me
giải giúp mik bt này vs mn!
1)\(\left\{{}\begin{matrix}2x^2+y^2+x=3\left(xy+1\right)+2y\\\dfrac{2}{3+\sqrt{2x-y}}+\dfrac{2}{3+\sqrt{4-5x}}=\dfrac{9}{2x-y+9}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\left(x+3y+1\right)\sqrt{2xy+2y}=y\left(3x+4y+3\right)\\\left(\sqrt{x+3}-\sqrt{2y-2}\right)\left(x-3+\sqrt{x^2+x+2y-4}\right)=4\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}\sqrt{2x-3}=\left(y^2+2011\right)\left(5-y\right)+\sqrt{y}\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)
1,\(\left\{{}\begin{matrix}x-y^2+1=0\\\sqrt{y^2+3}+x=2\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}x^4+2x^3y+x^2y^2=2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)
3,\(\left\{{}\begin{matrix}xy+x-2=0\\2x^3-x^2y+x^2+y^2-2xy-y=0\end{matrix}\right.\)
hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}\frac{x+1}{y-1}=5\\3\left(2x-2\right)-4\left(3x+4\right)=5\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}2x+y=4\\\left|x-2y\right|=3\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{2x}{x+1}+\frac{y}{y+1}=3\\\frac{x}{x+1}-\frac{3y}{y+1}=-1\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}y-\left|x\right|=1\\2x-y=1\end{matrix}\right.\)
10 , \(\left\{{}\begin{matrix}\sqrt{x+3y}=\sqrt{3x-1}\\5x-y=9\end{matrix}\right.\)
