Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
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 hệ phương trình :
a) \(\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.\)
Gỉai hệ phương trình đối xứng sau:
a.\(\left\{{}\begin{matrix}x^2+y^2-x-2y=19\\xy\left(x-1\right)\left(y-2\right)=20\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)+8xy=0\\\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}=-\dfrac{1}{4}\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}x^3-3x^2-9x+22=y^3+3y^2-9y\\x^2+y^2-x+y=\dfrac{1}{2}\end{matrix}\right.\)
giả các hệ phương trình sau :
a) \(\left\{{}\begin{matrix}\dfrac{-3}{x-y+1}+\dfrac{1}{x +y-2}=12\\\dfrac{2}{x-y+1}-\dfrac{3}{x+y-2}=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+2\left(y^2+2y\right)=10\\3x^2-\left(y^2+2y\right)=9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x-1}}-\dfrac{5}{\sqrt{y+2}}=\dfrac{9}{2}\\\dfrac{3}{\sqrt{x-1}}+\dfrac{2}{\sqrt{y+2}}=4\end{matrix}\right.\)
giải hệ pt:
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
Giải hệ phương trình:
a) \(\left\{{}\begin{matrix}2x-y=3\\x+2y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{3}{2}x-y=\dfrac{1}{2}\\3x-2y=1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=\sqrt{2}\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)
Giúp mình với!!!
GIẢI HPT
\(_{\left\{{}\begin{matrix}2\left(x+1\right)-\dfrac{y}{4}=10\\\dfrac{2x+3}{3-4x}=\dfrac{x-y+5}{6-2x+2y}\end{matrix}\right.\)
Giải hệ phương trình
a)\(\left\{{}\begin{matrix}\dfrac{2x-1}{x+2}-\dfrac{5}{y-1}=-\dfrac{14}{3}\\\dfrac{3}{x+2}+\dfrac{\left(2y+3\right)}{y-1}=8\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{2x}{1-x}+\dfrac{3}{y+2}=-\dfrac{2}{5}\\\dfrac{5}{1-x}-\dfrac{4y}{y+2}=\dfrac{1}{10}\end{matrix}\right.\)
