\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(1+x^2\right)^2\left(1+\dfrac{1}{y^4}\right)=8\\\left(1+y^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\end{matrix}\right.\)
Em cảm ơn ạ !!!
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)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)
giải hệ phương trình
a,\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\dfrac{4}{x+2}-\dfrac{1}{x-2y}=1\\\dfrac{20}{x+2y}+\dfrac{3}{x-2y}=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-2\right|=2\\\left|x-1\right|+y=3\end{matrix}\right.\)
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?
a)\(\left\{{}\begin{matrix}2\left|x-6\right|+3\left|y-1\right|=5\\5\left|x-6\right|-4\left|y+1\right|=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2\left|x+y\right|-\left|x-y\right|=9\\3\left|x+y\right|+2\left|x-y\right|+17\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}4\left|x+y\right|+3\left|x-y\right|=8\\3\left|x+y\right|-5\left|x-y\right|=6\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}x^2-xy=24\\2x-3y=1\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}3x-4y+1=0\\xy=3\left(x+y\right)-9\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}2x+3y=5\\3x^2-y^2+2y=4\end{matrix}\right.\)
giải hệ phương trình
1, \(\left\{{}\begin{matrix}2x^2+3y=17\\3x^2-2y=6\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-1\right|=2\\4\left|x-1\right|+3\left|y-1\right|=7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=2\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}x+y=2\\\left|2x-3y\right|=1\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}2x-y=1\\\left|x-y\right|=\left|2y-1\right|\end{matrix}\right.\)
6,\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+6\right)=xy\\\left(x+2\right)\left(y-2\right)=xy\end{matrix}\right.\)
7 , \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}4x^2-5\left(y+1\right)=\left(2x-3\right)^2\\3\left(7x+2\right)=5\left(2y-1\right)-3x\end{matrix}\right.\)
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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
GHPT: \(\left\{{}\begin{matrix}2x+\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{16}{3}\\2\left(x^2+y^2\right)+\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)
Giải hệ
a) \(\left\{{}\begin{matrix}2x^2-5xy-y^2=1\\y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^3+1=2\left(x^2-x+y\right)\\y^3+1=2\left(y^2-y+x\right)\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^2-2y^2=1\\2y^2-3z^2=1\\xy+yz+zx=1\end{matrix}\right.\left(x,y,z\in R\right)}\)
