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

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\3m-1-y-4y=-2m-6\end{matrix}\right.\)

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

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

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

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

Vậy hpt trên có nghiệm duy nhất \(\left\{{}\begin{matrix}x=m-1\\y=m+1\end{matrix}\right.\)

Ta có: y = x² \(\Leftrightarrow\) m + 1 = (m - 1)² \(\Leftrightarrow\) m + 1 = m² - 2m + 1

\(\Leftrightarrow\) m² - 3m = 0

\(\Leftrightarrow\) m(m - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=0\\m-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=0\\m=3\end{matrix}\right.\)

Vậy m = 0; m = 3 thì hpt trên có nghiệm duy nhất và thỏa mãn y = x²

Chúc bn học tốt!

a: Vì \(\dfrac{1}{2}\ne\dfrac{-2}{1}=-2\)

nên hệ phương trình luôn có nghiệm duy nhất

b: \(\left\{{}\begin{matrix}x-2y=4-m\\2x+y=8m+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2y=4-m\\4x+2y=16m+6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+4x=4-m+16m+6=15m+10\\2x+y=8m+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=15m+10\\y=8m+3-2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3m+2\\y=8m+3-6m-4=2m-1\end{matrix}\right.\)

Đặt \(A=x^2+y^2\)

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

\(=9m^2+12m+4+4m^2-4m+1\)

\(=13m^2+8m+5\)

\(=13\left(m^2+\dfrac{8}{13}m+\dfrac{5}{13}\right)\)

\(=13\left(m^2+2\cdot m\cdot\dfrac{4}{13}+\dfrac{16}{169}+\dfrac{49}{169}\right)\)

\(=13\left(m+\dfrac{4}{13}\right)^2+\dfrac{49}{13}>=\dfrac{49}{13}\forall m\)

Dấu '=' xảy ra khi \(m+\dfrac{4}{13}=0\)

=>\(m=-\dfrac{4}{13}\)

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 )

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

=>\(m^2\ne4\)

=>\(m\notin\left\{2;-2\right\}\)

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

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

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

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

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

=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\x\cdot\dfrac{4-m^2}{2}=-m^2-\dfrac{5}{2}m+9=\dfrac{-2m^2-5m+18}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{-2m^2-5m+18}{4-m^2}=\dfrac{2m^2+5m-18}{m^2-4}\\y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}x=\dfrac{2m+9}{m+2}\\y=\dfrac{2m^2+10m+2-2m^2-4m}{2\left(m+2\right)}=\dfrac{6m+2}{2\left(m+2\right)}=\dfrac{3m+1}{m+2}\end{matrix}\right.\)

Để x,y nguyên thì \(\left\{{}\begin{matrix}2m+9⋮m+2\\3m+1⋮m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m+4+5⋮m+2\\3m+6-5⋮m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5⋮m+2\\-5⋮m+2\end{matrix}\right.\)

=>\(5⋮m+2\)

=>\(m+2\in\left\{1;-1;5;-5\right\}\)

=>\(m\in\left\{-1;-3;3;-7\right\}\)

lấy (1) + 2.(2) 

sẽ có x = 2m-1 

thay vào (1) sẽ ra y = 2-m 

thay x và y vừa tìm được vào phần thỏa mãn sẽ có 2 nghiệm m = -1 hoặc m = \(\dfrac{3}{2}\) rồi thay vào tìm x và y theo 2 trường hợp 

trường hợp 1: m = -1 

thì ta tìm được x = -3 và y = 3 

trường hợp 2: m= \(\dfrac{3}{2}\)

x = 2 

y = \(\dfrac{1}{2}\)  

( mình chỉ bạn cách làm thôi nên hk có trình bày rõ bạn trình bày lại nhé)

=>2x-4y=8m-10 và 2x+y=3m

=>-5y=5m-10 và 2x+y=3m

=>y=-m+2 và 2x=3m+m-2=4m-2

=>y=-m+2 và x=2m-1

2/x-1/y=-1

=>\(\dfrac{2}{2m-1}+\dfrac{1}{m-2}=-1\)

=>\(\dfrac{2m-4+2m-1}{\left(m-2\right)\left(2m-1\right)}=-1\)

=>-(2m^2-m-4m+2)=4m-5

=>2m^2-5m+2=-4m+5

=>2m^2+m-3=0

=>(2m+3)(m-1)=0

=>m=1 hoặc m=-3/2