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a: Khi m=2 thì hệ sẽ là;
2x-y=4 và x-2y=3
=>x=5/3 và y=-2/3
b: mx-y=2m và x-my=m+1
=>x=my+m+1 và m(my+m+1)-y=2m
=>m^2y+m^2+m-y-2m=0
=>y(m^2-1)=-m^2+m
Để phương trình có nghiệm duy nhất thì m^2-1<>0
=>m<>1; m<>-1
=>y=(-m^2+m)/(m^2-1)=(-m)/m+1
x=my+m+1
\(=\dfrac{-m^2+m^2+2m+1}{m+1}=\dfrac{2m+1}{m+1}\)
x^2-y^2=5/2
=>\(\left(\dfrac{2m+1}{m+1}\right)^2-\left(-\dfrac{m}{m+1}\right)^2=\dfrac{5}{2}\)
=>\(\dfrac{4m^2+4m+1-m^2}{\left(m+1\right)^2}=\dfrac{5}{2}\)
=>2(3m^2+4m+1)=5(m^2+2m+1)
=>6m^2+8m+2-5m^2-10m-5=0
=>m^2-2m-3=0
=>(m-3)(m+1)=0
=>m=3
\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y+x+2y=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)
\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy ...
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
Khi \(m\notin\left\{1;-1\right\}\) thì \(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y-2m=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-m^2+1\right)=-m^2+m\\x=m+1-my\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m^2-m}{m^2-1}=\dfrac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{m}{m+1}\\x=m+1-\dfrac{m^2}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m}{m+1}\\x=\dfrac{\left(m+1\right)^2-m^2}{m+1}=\dfrac{2m+1}{m+1}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>=2\\y>=1\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}>=2\\\dfrac{m}{m+1}>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2\left(m+1\right)}{m+1}>=0\\\dfrac{m-m-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2m-2}{m+1}>=0\\\dfrac{-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{m+1}>=0\\-\dfrac{1}{m+1}>=0\end{matrix}\right.\Leftrightarrow m+1< 0\)
=>m<-1
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...
Để hệ có nghiệm duy nhất thì \(\dfrac{m}{2m}\ne\dfrac{1}{3}\)
=>\(\dfrac{1}{2}\ne\dfrac{1}{3}\)(luôn đúng)
\(\left\{{}\begin{matrix}mx+y=5\\2mx+3y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2mx+2y=10\\2mx+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=4\\mx+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-4\\mx=5-y=5-\left(-4\right)=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-4\\x=\dfrac{9}{m}\end{matrix}\right.\)
\(\left(2m-1\right)\cdot x+\left(m+1\right)\cdot y=m\)
=>\(\dfrac{9}{m}\left(2m-1\right)+\left(m+1\right)\cdot\left(-4\right)=m\)
=>\(\dfrac{9\left(2m-1\right)}{m}=m+4m+4=5m+4\)
=>m(5m+4)=18m-9
=>\(5m^2-14m+9=0\)
=>(m-1)(5m-9)=0
=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{9}{5}\end{matrix}\right.\)
a. Thay m = 1 ta được
\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
b, Để hpt có nghiệm duy nhất khi \(\dfrac{1}{2}\ne-\dfrac{2}{3}\)*luôn đúng*
\(\left\{{}\begin{matrix}2x+4y=2m+6\\2x-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=m+6\\x=m+3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m+6}{7}\\x=m+3-2\dfrac{m+6}{7}\end{matrix}\right.\)
\(\Leftrightarrow x=m+3-\dfrac{2m+12}{7}=\dfrac{7m+21-2m-12}{7}=\dfrac{5m+9}{7}\)
Ta có : \(\dfrac{m+6}{7}+\dfrac{5m+9}{7}=-3\Rightarrow6m+15=-21\Leftrightarrow m=-6\)
\(\left\{{}\begin{matrix}x+2y=m+3\\2x-3y=m\end{matrix}\right.\)
\(a,Khi.m=1\Rightarrow\left\{{}\begin{matrix}x+2y=1+3\\2x-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\2\left(4-2y\right)-3y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4-2y\\8-4y-3y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4-2y\\7y=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\rightarrow\left(x,y\right)=\left(2,1\right)\)
\(b,\left\{{}\begin{matrix}x+2y=m+3\\2x-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2m+6\left(1\right)\\2x-3y=m\left(2\right)\end{matrix}\right.\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}7y=m+6\\x+2y=m+3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+9}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\Rightarrow\) HPT có no duy nhất
\(\left(x,y\right)=\left(\dfrac{5m+9}{7};\dfrac{m+6}{7}\right)\)
\(x+y=-3\)
\(\dfrac{5m+9}{7}+\dfrac{m+6}{7}=-3\)
\(\Leftrightarrow5m+9+m+6=-21\)
\(\Leftrightarrow6m=-36\Rightarrow m=-6\)
Với m = -6 thì hệ pt có no duy nhất TM x + y = -3
Kết hợp điều kiện đề bài và pt thứ 2 của hệ ta được:
\(\left\{{}\begin{matrix}x-y=-6\\2x+y=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=7\end{matrix}\right.\)
Thế vào pt đầu:
\(m.1+2.7=18\Rightarrow m=4\)
mình giải tắt nhé vì mình không giỏi dùng công thức. Thông cảm nha.
1.
\(\left\{{}\begin{matrix}3x-y=2m+3\\x+y=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m}{4}+1\\y=\dfrac{-5m}{4}\end{matrix}\right.\)
vậy phương trình có nghiệm duy nhất là \(\left(\dfrac{m}{4}+1;\dfrac{-5m}{4}\right)\)
Thay vào đẳng thức ta được:
\(\left(\dfrac{m}{4}+1\right)^2+\left(\dfrac{-5m}{4}\right)^2=5\\ \Leftrightarrow x=\)
k sao đâu bạn mình cảm ơn ạ