a) m = 3 thì hệ trở thành \(\hept{\begin{cases}3x+y=3\\2x-y=7\end{cases}}\Leftrightarrow\hept{\begin{cases}6x+2y=6\left(1\right)\\6x-3y=21\left(2\right)\end{cases}}\)

\(\left(1\right)-\left(2\right)\Leftrightarrow5y=-15\Leftrightarrow y=-3\)

Từ đó suy ra \(x=2\)

Vậy với m = 3 thì hệ có 1 nghiệm (2;-3)

b) HPT không thể có nghiệm (3;1)

c) HPT có nghiệm (4;1) thì \(4m+1=3\Leftrightarrow m=\frac{1}{2}\)

a/ Xét pt : \(\left\{{}\begin{matrix}mx-y=1\\\dfrac{x}{2}-\dfrac{y}{2}=335\end{matrix}\right.\)

 Khi \(m=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-y=670\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-669\\y=-1339\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}mx-y=1\\x-y=670\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-670\\mx-\left(x-670\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-670\\x\left(m-1\right)=-669\end{matrix}\right.\)

Để pt có nghiệm duy nhất \(\Leftrightarrow m\ne1\)

Vậy...

a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=6\\-5x+8y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15x-10y=30\\-15x+24y=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=21\\3x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-\dfrac{3}{4}\end{matrix}\right.\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+3\sqrt{y-3}=18\\9\sqrt{x}-3\sqrt{y-3}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y-3}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}46a+46b+14b=20,5\\a+b=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}46\left(a+b\right)+14b=20,5\\a+b=0,4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}46.0,4+14b=20,5\\a+b=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}14b=2,1\\a+b=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=0,15\\a+0,15=0,4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=0,15\\a=0,25\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x+y=3\\3x+2y=2\end{matrix}\right.\)

Từ `(1)` : `2x+y=3=> y=3-2x` thế vào `(2)`

Ta được :

`3x+2(3-2x)=2`

`<=> 3x+6-4x=2`

`<=> -x =-4`

`<=>x=4`

`=> y= 3-2.4`

`<=> y= 3-6`

`<=>y=-3`

Vậy nghiệm của hệ `(x,y)` là `(4;-3)`

Để hệ vô nghiệm thì

\(\dfrac{2m-1}{m+3}\ne\dfrac{5}{1}\\ \Leftrightarrow2m-1\ne5\left(m+3\right)\\ \Leftrightarrow m\ne-\dfrac{16}{3}\)

\(\dfrac{2m-1}{m+3}=\dfrac{3}{-4}\left(m\ne-3\right)\\ \Leftrightarrow-4\left(2m-1\right)=3\left(m+3\right)\\ \Leftrightarrow m=-\dfrac{5}{11}\left(tm\right)\)