`{(3x - 2y = 10),(x-2/3 y = (3)(1/3)):}`
`<=> {(3x - 2y = 10),(x-2/3 y = 10/3):}`
`<=> {(3x - 2y = 10),(3x-2 y = 10):}`
`<=> {(0x=0),(3x-2 y = 10):}`
`<=> {(0x=0),(y = (3x-10)/2):}`
`<=> {(x∈ R),(y = (3x-10)/2):}`
Vậy pt có vô số nghiệm
giải các hpt sau: a)\(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{3x}{4}+\dfrac{2y}{5}=2,3\\x-\dfrac{3y}{5}=0,8\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)cíu zới
Giải các hệ phương trình sau bằng phương pháp thế:
a)\(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
Giải HPT bằng phương pháp đặt ẩn phụ
\(\left\{{}\begin{matrix}\dfrac{6}{x+y}-\dfrac{3}{x-2y}=3\\\dfrac{1}{x+y}+\dfrac{7}{x-2y}=2\end{matrix}\right.\)
Giải hệ pt sau = phương pháp thế:
a, \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}2\left(x+1\right)-3y=-10\\3x+2y+5=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x+1}{2}-\dfrac{y-2}{3}=1\\4x+3y=1\end{matrix}\right.\)
Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
giải hệ phương trình
1)\(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\) 2)\(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\) 3)\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}-x+3y=16\\2x+y=3\end{matrix}\right.\) 5)\(\left\{{}\begin{matrix}\dfrac{-3}{x-y}+\dfrac{5}{2x+y}=-2\\\dfrac{4}{x-y}-\dfrac{10}{2x+y}=2\end{matrix}\right.\) 6)\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
giải hpt
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)
giải hpt sau
\(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)
\(6.\left\{{}\begin{matrix}x+2y=5\\3x-y=1\end{matrix}\right.\)
\(7.\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y-3\right)=xy-3\end{matrix}\right.\)
\(8.\left\{{}\begin{matrix}\dfrac{1}{x+1}-\dfrac{3}{y-1}=-1\\\dfrac{2}{x+1}+\dfrac{4}{y-1}=3\end{matrix}\right.\)
