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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
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 \(\left\{{}\begin{matrix}2m+9⋮m+2\\3m+1⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+4+5⋮m+2\\3m+6-5⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5⋮m+2\\-5⋮m+2\end{matrix}\right.\)
=>\(5⋮m+2\)
=>\(m+2\in\left\{1;-1;5;-5\right\}\)
=>\(m\in\left\{-1;-3;3;-7\right\}\)
Hệ có nghiệm duy nhất khi: \(\dfrac{3}{1}\ne\dfrac{m}{-2}\Rightarrow m\ne-6\)
Khi đó ta có:
\(\left\{{}\begin{matrix}3x+my=5\\x-2y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x+2my=10\\mx-2my=3m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+6\right)x=3m+10\\y=\dfrac{x-3}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{x-3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{-4}{m+6}\end{matrix}\right.\)
\(2x+y=1\Rightarrow\dfrac{2\left(3m+10\right)}{m+6}+\dfrac{-4}{m+6}=1\)
\(\Leftrightarrow\dfrac{6m+16}{m+6}=1\)
\(\Rightarrow6m+16=m+6\)
\(\Rightarrow m=-2\)
Hệ \(\Leftrightarrow\left\{{}\begin{matrix}x=3m-my\\mx-y=m^2-2\end{matrix}\right.\)
\(\Rightarrow m\left(3m-my\right)-y=m^2-2\)
\(\Leftrightarrow2m^2+2=y\left(1+m^2\right)\)
\(\Leftrightarrow y=\dfrac{2m^2+2}{1+m^2}=2\)
\(\Rightarrow x=3m-2m=m\)
Có \(x^2-2x-y>0\Leftrightarrow m^2-2m-2>0\)
\(\Leftrightarrow\left(m-1-\sqrt{3}\right)\left(m-1+\sqrt{3}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m>1+\sqrt{3}\\m< 1-\sqrt{3}\end{matrix}\right.\)
Vậy...
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 \(m+\dfrac{4}{13}=0\)
=>\(m=-\dfrac{4}{13}\)
\(\left\{{}\begin{matrix}x+my=3\\x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)y=2\\x=1-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{m-2}\\x=1-\dfrac{4}{m-2}=\dfrac{m-6}{m-2}\end{matrix}\right.\)
a, Ta có x < 0 ; y > 0
\(x< 0\Rightarrow\dfrac{m-6}{m-2}< 0\)
Ta có : m - 2 > m - 6
\(\left\{{}\begin{matrix}m-2>0\\m-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>2\\m< 6\end{matrix}\right.\Leftrightarrow2< m< 6\)
\(y>0\Leftrightarrow\dfrac{2}{m-2}>0\Rightarrow m>2\)
Vậy 2 < m < 6
b, \(x-2y=3\Rightarrow\dfrac{m-6}{m-2}-\dfrac{4}{m-2}=3\Leftrightarrow\dfrac{m-10}{m-2}=3\)
\(\Rightarrow m-10=3m-6\Leftrightarrow2m=-4\Leftrightarrow m=-2\)
\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y+x+2y=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)
\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy ...
\(\left\{{}\begin{matrix}2x+y=3m-1\\x-2y=-m-3\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\\dfrac{3m-1-y}{2}-2y=-m-3\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\3m-1-y-4y=-2m-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\5y=5m+5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\y=m+1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-m-1}{2}\\y=m+1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m-1\\y=m+1\end{matrix}\right.\)
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
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