giải hệ phương trình

a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)

c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)

Giải các hệ phương trình sau : 

a, \(\left\{{}\begin{matrix}x^2+xy=y^2+1\\3x+y=y^2+3\end{matrix}\right.\)

b,\(\left\{{}\begin{matrix}x^2-y^2=4x-2y-3\\x^2+y^2=5\end{matrix}\right.\)

c, \(\left\{{}\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)

d,\(\left\{{}\begin{matrix}2\left(y+z\right)=yz\\xy+yz+zx=108\\xyz=180\end{matrix}\right.\)

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?

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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

1, Giải các hệ phương trình sau

a, \(\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=26\\x+y=6\end{matrix}\right.\)

b,\(\left\{{}\begin{matrix}2x^2+x-y=0\\xy+3y-5x=7\end{matrix}\right.\)

c, \(\left\{{}\begin{matrix}\left(x-1\right)^2=1-y\\\left(x^2-y\right)^2=2xy\left(1+x\right)\end{matrix}\right.\)

d, \(\left\{{}\begin{matrix}x^2y+y^2x=2\\x^3+y^3+6=8x^2y^2\end{matrix}\right.\)