Để hệ có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)

=>\(2m\ne m-1\)

=>\(m\ne-1\)(1)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(m-1\right)-2mx+m^2+5m-3m+1=0\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(-m-1\right)+m^2+2m+1=0\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(m+1\right)=\left(m+1\right)^2\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\end{matrix}\right.\)

\(x^2-y^2< 4\)

=>\(\left(m+1\right)^2-\left(m-3\right)^2< 4\)

=>\(m^2+2m+1-m^2+6m-9< 4\)

=>8m-8<4

=>8m<12

=>\(m< \dfrac{3}{2}\)

Kết hợp (1), ta được: \(\left\{{}\begin{matrix}m< \dfrac{3}{2}\\m\ne-1\end{matrix}\right.\)

Để hệ có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)

=>\(2m\ne m-1\)

=>\(m\ne-1\)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(m-1\right)-2mx+m^2+5m=3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(m-1-2m\right)=-m^2-5m+3m-1=-m^2-2m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(-m-1\right)=-\left(m+1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\end{matrix}\right.\)

\(x^2-y^2< 4\)

=>\(\left(m+1\right)^2-\left(m-3\right)^2< 4\)

=>\(m^2+2m+1-m^2+6m-9< 4\)

=>8m-8<4

=>8m<12

=>\(m< \dfrac{3}{2}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}m< \dfrac{3}{2}\\m\ne-1\end{matrix}\right.\)

a) Thay m=2 vào hệ phương trình, ta được: 

\(\left\{{}\begin{matrix}x-2y=5\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=10\\2x-y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3y=3\\x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=5+2y=5+2\cdot\left(-1\right)=3\end{matrix}\right.\)

Vậy: Khi m=2 thì hệ phương trình có nghiệm duy nhất là (x,y)=(3;-1)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2mx-my=m^2+5m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\\left(m+1\right)x=m^2+2m+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\\left(m+1\right)x=\left(m+1\right)^2\end{matrix}\right.\)

Pt có nghiệm duy nhất \(\Leftrightarrow m\ne-1\)

Khi đó: \(\left\{{}\begin{matrix}x=m+1\\y=m-3\end{matrix}\right.\)

\(x^2-y^2=4\Leftrightarrow\left(m+1\right)^2-\left(m-3\right)^2=4\)

\(\Leftrightarrow8m=12\Rightarrow m=\dfrac{3}{2}\)

`x+my=m+1=>x=m+1-my` thế vào dưới

`=>m(m+1-my)+y-3m+1=0`

`<=>m^2+m-my^2+y-3m-1`

`=>y(1-m^2)=2m-1-m^2`

Hệ có no duy nhất

`=>1-m^2 ne 0=>m ne +-1`

`=>y=(-1+2m-m^2)/(1-m^2)=(m-1)/(m+1)`

`=>x=m+1-my=((m+1)^2-m(m-1))/(m+1)=(3m+1)/(m+1)`

`=>xy=((3m+1)(m-1))/(m+1)^2=(3m^2-2m-1)/(m+1)^2`

Xét `xy+1`

`=(3m^2-2m-1+m^2+2m+1)/(m+1)^2=(4m^2)/(m+1)^2`

`=>xy+1>=0=>xy>=-1`

Dấu "=" xảy ra khi `m=0`

em tham khảo câu e cuối ấy , đó là câu a của e á:

Vì \(\dfrac{3}{1}\ne\dfrac{-1}{2}\)

nên hệ luôn có nghiệm duy nhất

\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-y=2m-1\\3x+6y=9m+6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=2m-1-9m-6=-7m-7\\x+2y=3m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=m+1\\x=3m+2-2m-2=m\end{matrix}\right.\)

\(y-\sqrt{x}=1\)

=>\(m+1-\sqrt{m}=1\)

=>\(m-\sqrt{m}=0\)

=>\(\sqrt{m}\left(\sqrt{m}-1\right)=0\)

=>\(\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\)

Bài 1.

\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)

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

\(x_0^2+y_0^2=9m\)

\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)

\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)

\(\Leftrightarrow2m^2-7m+5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )

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

\(\Leftrightarrow\left\{{}\begin{matrix}3x=3m\\2x-y=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m-1\end{matrix}\right.\)

Theo đề: \(x^2-2y-1=0\)

\(\Leftrightarrow m^2-2\left(m-1\right)-1=0\)

\(\Leftrightarrow m^2-2m+1=0\)

\(\Leftrightarrow\left(m-1\right)^2=0\Leftrightarrow m=1\).

Vậy: \(m=1.\)