a, Ta có : \(\dfrac{4}{6}=-\dfrac{2}{-3}\ne\dfrac{5}{5}=1\)
vậy hpt vô nghiệm
b, Ta có \(\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{5}{10}\)-> hệ pt có vô số nghiệm
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}2x+5y=5\\3x-5y=-30\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}4x-3y=-5\\3x+2y=-8\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x+3y=9\\4x-2y=-2\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}5x-4y=32\\6x+2y=18\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}2x-3y+5=0\\3x+5y-21=0\end{matrix}\right.\) f) \(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x\sqrt{2}+y=5\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}5x+4y=-3\\3x+2y=11\end{matrix}\right.\) h) \(\left\{{}\begin{matrix}2x-4y=12\\5x+3y=17\end{matrix}\right.\)
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 hệ phương trình
a
\(\left\{{}\begin{matrix}x+y=1\\x-y=-5\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}2x+2y=5\\x-2y=1\end{matrix}\right.\)
c.
\(\left\{{}\begin{matrix}2x+3y=5\\3x-2y=1\end{matrix}\right.\)
1)Nghiệm duy nhất
\(\left\{{}\begin{matrix}mx+2y=7\\2x+3y=5\end{matrix}\right.\)
1)Vô nghiệm
\(\left\{{}\begin{matrix}2x-y=m\\-4x+2y=4\end{matrix}\right.\)
a)Tìm nghiệm duy nhất
\(\left\{{}\begin{matrix}mx+2y=7\\2x+3y=5\end{matrix}\right.\)
b) Vô nghiệm
\(\left\{{}\begin{matrix}2x-y=m\\-4x+2y=4\end{matrix}\right.\)
giải hpt sau
\(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3x+y=3\\2x-y=7\end{matrix}\right.\)
..
\(\left\{{}\begin{matrix}2x+5y=8\\2x-3y=0\end{matrix}\right.\)
..
\(\left\{{}\begin{matrix}4x+3y=6\\2x+y=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2x+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
