1)Nghiệm duy nhất
\(\left\{{}\begin{matrix}mx+2y=7\\2x+3y=5\end{matrix}\right.\)
1)Vô nghiệm
\(\left\{{}\begin{matrix}2x-y=m\\-4x+2y=4\end{matrix}\right.\)
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8 tháng 1 2021
1)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;2\right)\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
3)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
4)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
8 tháng 1 2021
1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)
=> Hệ có vô số nghiệm.
3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
21 tháng 3 2020
1) \(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\8x+3\left(2-4x\right)=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\)
2) 2 pt 3 ẩn không giải được.
3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+1}{2}\\-4\cdot\frac{3y+1}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-3y+5}{2}\\5\cdot\frac{-3y+5}{2}-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}x+4y=2\\3x+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2-4y\\3\left(2-4y\right)+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1}{5}\\x=\frac{6}{5}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-x-2\\-2x-3\left(-x-2\right)=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
18 tháng 1 2021
Với m = 0 ta có hpt \(\left\{{}\begin{matrix}2y=1\\2x=-1\end{matrix}\right.\). HPT này không có nghiệm nguyên.
Xét \(m\neq 0\).
Để hpt có nghiệm duy nhất thì: \(\dfrac{m}{2}\ne\dfrac{2}{m}\Leftrightarrow m\ne\pm2\).
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2mx+4y=2m+2\\2mx+m^2y=2m^2-m\end{matrix}\right.\Rightarrow\left(m^2-4\right)y=2m^2-3m-2\).
\(\Rightarrow y=\dfrac{2m^2-3m-2}{m^2-4}=\dfrac{2m+1}{m+2}\).
Từ đó ta có \(x=\dfrac{m+1-\dfrac{2\left(2m+1\right)}{m+2}}{m}=\dfrac{m^2+3m+2-4m-2}{m\left(m+2\right)}=\dfrac{m^2-m}{m\left(m+2\right)}=\dfrac{m-1}{m+2}\).
Vậy m là các số sao cho \(\dfrac{2m+1}{m+2}\) là số nguyên (Do \(\dfrac{2m+1}{m+2}-\dfrac{m-1}{m+2}=1\) là số nguyên).
\(1;\left\{{}\begin{matrix}mx+2y=7\\2x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7-mx}{2}\\2x+\dfrac{3\left(7-mx\right)}{2}=5\left(1\right)\end{matrix}\right.\)
\(hệ\) \(pt\) \(có\) \(nghiệm\) \(duy\) \(nhất\Leftrightarrow\left(1\right)có\) \(ngo\) \(duy\) \(nhất\)
\(\left(1\right)\Leftrightarrow\dfrac{4x+3\left(7-mx\right)}{2}=5\Leftrightarrow4x+21-3mx=10\Leftrightarrow x\left(4-3m\right)=-11\)
\(với:m\ne\dfrac{4}{3}\) \(thì\) \(hpt\) \(có\) \(ngo\) \(duy-nhất\left(x;y\right)=\left\{\dfrac{-11}{4-3m};\dfrac{7-m\left(\dfrac{-11}{4-3m}\right)}{2}\right\}\)
\(2,\left\{{}\begin{matrix}2x-y=m\\-4x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x-m\\-4x+2\left(2x-m\right)=4\left(1\right)\end{matrix}\right.\)
hệ pt vô nghiệm khi (1) vô nghiệm
(1)\(\Leftrightarrow-4x+4x-2m=4\Leftrightarrow m=-2\Rightarrow m=-2\)
thì hệ pt có vô số nghiệm
\(\Rightarrow m\ne-2\) thì hpt vô nghiệm