a. Bạn tự giải

b. \(\left\{{}\begin{matrix}6x+2my=2m\\\left(m^2-m\right)x+2my=m^2-m\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6x+2my=2m\\\left(m^2-m-6\right)x=m^2-3m\end{matrix}\right.\)

Hệ có nghiệm duy nhất khi \(m^2-m-6\ne0\Rightarrow m\ne\left\{-2;3\right\}\)

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

\(x+y^2=1\Leftrightarrow\dfrac{m}{m+2}+\left(\dfrac{m-1}{m+2}\right)^2=1\)

\(\Leftrightarrow m^2-4m-3=0\)

\(\Leftrightarrow...\)

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 )

mọi người ơi giúp mình vs mai ktra r

Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{4}\)

=>\(m^2\ne4\)

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

Ta có: \(\left\{{}\begin{matrix}x+my=1\\mx+4y=2\end{matrix}\right.\)

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

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

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

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

x+y>-5

=>\(\dfrac{2}{m+2}+\dfrac{1}{m+2}>-5\)

=>\(\dfrac{3}{m+2}+5>0\)

=>\(\dfrac{3+5m+10}{m+2}>0\)

=>\(\dfrac{5m+13}{m+2}>0\)

TH1: \(\left\{{}\begin{matrix}5m+13>0\\m+2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>-\dfrac{13}{5}\\m>-2\end{matrix}\right.\)

=>\(m>-2\)

TH2: \(\left\{{}\begin{matrix}5m+13< 0\\m+2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< -\dfrac{13}{5}\\m< -2\end{matrix}\right.\)

=>\(m< -\dfrac{13}{5}\)

Vậy: \(\left[{}\begin{matrix}m< -\dfrac{13}{5}\\\left\{{}\begin{matrix}m>-2\\m\ne2\end{matrix}\right.\end{matrix}\right.\)

Lời giải:

$x+my=2\Rightarrow x=2-my$. Thay vào PT(2):

$m(2-my)-2y=1$

$\Leftrightarrow 2m-y(m^2+2)=1$

$\Leftrightarrow y=\frac{2m-1}{m^2+2}$

$x=2-my=2-\frac{2m^2-m}{m^2+2}=\frac{m+4}{m^2+2}$

Vậy hpt có nghiệm $(x,y)=(\frac{m+4}{m^2+2}; \frac{2m-1}{m^2+2})$

Để $x<0; y>0$

$\Leftrightarrow \frac{m+4}{m^2+2}<0$ và $\frac{2m-1}{m^2+2}>0$

$\Leftrightarrow m+4<0$ và $2m-1>0$ (do $m^2+2>0$)

$\Leftrightarrow m< -4$ và $m> \frac{1}{2}$  (vô lý)

Do đó không tồn tại $m$ thỏa mãn đề.

Hệ có nghiệm duy nhất khi: \(\dfrac{1}{m}\ne\dfrac{m}{-2}\Rightarrow m^2\ne-2\) (luôn đúng)

\(\Rightarrow\) Hệ luôn có nghiệm duy nhất với mọi m

Khi đó: \(\left\{{}\begin{matrix}x+my=2\\mx-2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+2my=4\\m^2x-2my=m\end{matrix}\right.\)

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

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

Nghiệm hệ thỏa mãn x<0, y<0 \(\Rightarrow\left\{{}\begin{matrix}\dfrac{m+4}{m^2+2}< 0\\\dfrac{4m-2}{2\left(m^2+2\right)}< 0\end{matrix}\right.\) (1)

Do \(m^2+2>0;\forall m\) nên (1) tương đương:

\(\left\{{}\begin{matrix}m+4< 0\\4m-2< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m< -4\\m< \dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow m< -4\)

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{m}{3}\\-2y=\dfrac{10}{3}m+4\end{matrix}\right.\)

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

Để \(x^2+y^2=10\)

\(\Leftrightarrow\left(\dfrac{-m}{3}\right)^2+\left(\dfrac{-5x}{3}-2\right)^2=10\)

\(\Leftrightarrow\dfrac{m^2}{9}+\dfrac{25m^2}{9}+\dfrac{20m}{3}+4=10\)

\(\Leftrightarrow\dfrac{26m^2}{9}+\dfrac{20m}{3}-6=0\)

\(\Leftrightarrow\dfrac{26m^2}{9}+\dfrac{60m}{9}-\dfrac{54}{9}=0\)

\(\Leftrightarrow26m^2+60m-54=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-3\\m=\dfrac{9}{13}\end{matrix}\right.\)

a:

Để hệ có nghiệm duy nhất thì m/2<>-2/-m

=>m^2<>4

=>m<>2 và m<>-2