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a: Thay m=-1 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x-y=3\cdot\left(-1\right)=-3\\-x-y=\left(-1\right)^2-2=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2y=-6\\x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=y-3=3-3=0\end{matrix}\right.\)
a: Khi m=3 thì hệ phương trình sẽ là:
\(\left\{{}\begin{matrix}3x-y=2\\2x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x-3y=6\\2x+3y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11x=11\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3x-2=3-2=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}mx-y=2\\2x+my=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-2\\2x+m\left(mx-2\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\x\left(m^2+2\right)=5+2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-2\\x=\dfrac{2m+5}{m^2+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2m^2+5m}{m^2+2}-2=\dfrac{2m^2+5m-2m^2-4}{m^2+2}=\dfrac{5m-4}{m^2+2}\\x=\dfrac{2m+5}{m^2+2}\end{matrix}\right.\)
\(x+y=1-\dfrac{m^2}{m^2+2}\)
=>\(\dfrac{5m-4+2m+5}{m^2+2}=\dfrac{m^2+2-m^2}{m^2+2}=\dfrac{2}{m^2+2}\)
=>7m+1=2
=>7m=1
=>\(m=\dfrac{1}{7}\)
a: Thay m=-2 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x-2y=-2+1=-1\\-2x+y=3\cdot\left(-2\right)-1=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-4y=-2\\-2x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3y=-9\\x-2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x=2y-1=2\cdot3-1=5\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\y\left(-m^2+1\right)=3m-1-m^2-m=-m^2+2m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\y\left(m-1\right)\left(m+1\right)=\left(m-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=m+1-m\cdot\dfrac{m-1}{m+1}=\left(m+1\right)-\dfrac{m^2-m}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\end{matrix}\right.\)
\(x^2-y^2=4\)
=>\(\dfrac{\left(3m+1\right)^2-\left(m-1\right)^2}{\left(m+1\right)^2}=4\)
=>\(\dfrac{9m^2+6m+1-m^2+2m+1}{\left(m+1\right)^2}=4\)
=>\(8m^2+8m+2=4\left(m+1\right)^2\)
=>\(8m^2+8m+2-4m^2-8m-4=0\)
=>\(4m^2-2=0\)
=>\(m^2=\dfrac{1}{2}\)
=>\(m=\pm\dfrac{1}{\sqrt{2}}\)
Mình mạn phép sửa lại phương trình $2$ của bạn là $mx+3y=1$ nhé.
ĐK: $m\neq 0$
a) Khi $m=2,$ hệ phương trình là:
\(\left\{{}\begin{matrix}-4x+y=5\\2x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+y=5\\4x+6y=2\end{matrix}\right.\Rightarrow7y=7\Leftrightarrow y=1\Rightarrow x=-1\)
b) \(\left\{{}\begin{matrix}-2mx+y=5\\mx+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2mx+y=5\\2mx+6y=2\end{matrix}\right.\Rightarrow7y=7\Leftrightarrow y=1\Rightarrow x=-\dfrac{2}{m}\)
c) Do ta luôn có $y=1$ là số dương nên chỉ cần chọn $m$ sao cho:
\(x=-\dfrac{2}{m}>0\Leftrightarrow m< 0\)
d) \(x^2+y^2=1\Leftrightarrow\left(-\dfrac{2}{m}\right)^2+1^2=1\Leftrightarrow\dfrac{4}{m^2}=0\) (vô lý)
Vậy không tồn tại $m$ sao cho $x^2+y^2=1.$
a)Với m=2 thì hpt trở thành:
x-2y=5
2x-y=7
<=>
2x-4y=10
2x-y=7
<=>
-3y=3
2x-y=7
<=>
y=-1
x=3
b)\(\int^{\left(m-1\right)x-my=3m-1}_{2x-y=m+5}\Leftrightarrow\int^{x=\frac{3m+my-1}{m-1}}_{\frac{6m+2my-2}{m-1}-y=m+5}\Leftrightarrow\int^{x=\frac{3m+my-1}{m-1}}_{m^2+2m+my+y+3=0}\)
*m2+2m+my+y+3=0
<=>y.(m+1)=-m2-2m-3
*Với m=-1 =>PT vô nghiệm
*Với m khác -1 =>PT có nghiệm là: \(y=\frac{-m^2-2m-3}{m+1}=-m-1-\frac{2}{m+1}\)
bí tiếp
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
ai giải mk vs ạ