Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
\(\left\{{}\begin{matrix}mx-y=2\\3x+my=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\3x+m\left(mx-2\right)=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+xm^2-2m=5\\y=mx-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(m^2+3\right)=2m+5\\y=mx-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2m+5}{m^2+3}\\y=\frac{m\left(2m+5\right)}{m^2+3}-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2m+5}{m^2+3}\\y=\frac{5m-6}{m^2+3}\end{matrix}\right.\)
Khi đó: \(x+y=1-\frac{m^2}{m^2+3}\)
\(\Leftrightarrow\frac{2m+5}{m^2+3}+\frac{5m-6}{m^2+3}=1-\frac{m^2}{m^2+3}\)
\(\Leftrightarrow\frac{7m-4}{m^2+3}=0\)
\(\Leftrightarrow7m-4=0\)
\(\Leftrightarrow m=\frac{4}{7}\)
Vậy...
2.
\(\left\{{}\begin{matrix}2x+y=a+2\\x-y=a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=a+y\\2a+2y+y=a+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y=-a+2\\x=a+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-a+2}{3}\\x=a+\frac{-a+2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-a+2}{3}\\x=\frac{2a+2}{3}\end{matrix}\right.\)
\(x< y\Leftrightarrow\frac{2a+2}{3}< \frac{-a+2}{3}\)
\(\Leftrightarrow\frac{2a+2+a-2}{3}< 0\)
\(\Leftrightarrow\frac{3a}{3}< 0\)
\(\Leftrightarrow a< 0\)
Vậy...
1.Để đường thẳng \(y=\left(m-1\right)x+3\) song song với đường thẳng \(y=2x+1\)
thì \(m-1=2\Rightarrow m=3\)
2. a. Với \(m=-2\Rightarrow\)\(\hept{\begin{cases}-2x-2y=3\\3x-2y=4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=-\frac{17}{10}\end{cases}}\)
b. Với \(m=0\Rightarrow\hept{\begin{cases}-2y=3\\3x=4\end{cases}\Rightarrow\hept{\begin{cases}y=-\frac{3}{2}\\x=\frac{4}{3}\end{cases}\left(l\right)}}\)
Với \(m\ne0\Rightarrow\hept{\begin{cases}m^2x-2my=3m\\6x+2my=8\end{cases}\Rightarrow\left(m^2+6\right)x=3m+8}\)
\(\Rightarrow x=\frac{3m+8}{m^2+6}\)\(\Rightarrow y=\frac{mx-3}{2}=\frac{m\left(3m+8\right)-3\left(m^2+6\right)}{2\left(m^2+6\right)}=\frac{4m-9}{m^2+6}\)
Để \(x+y=5\Rightarrow\frac{3m+8}{m^2+6}+\frac{4m-9}{m^2+6}=5\Rightarrow7m-1=5m^2+30\)
\(\Rightarrow-5m^2+7m-31=0\)
Ta thấy phương trình vô nghiệm nên không tồn tại m để \(x+y=5\)
\(\int^{mx-y=2}_{3x+my=5}\Leftrightarrow\int^{y=mx-2}_{3x+m.\left(mx-2\right)=5}\Leftrightarrow\int^{y=mx-2}_{3x+m^2x-2m=5}\)
*3x+m2x-2m=5
<=>x.(3+m2)=5+2m
<=>x=\(\frac{5+2m}{3+m^2}\left(3+m^2>0\right)\)
\(\Rightarrow y=m.\left(\frac{5+2m}{3+m^2}\right)-2=\frac{5m+2m^2}{3+m^2}-\frac{6+2m^2}{3+m^2}=\frac{5m-6}{3+m^2}\)
=>\(x+y=\frac{5+2m}{3+m^2}+\frac{5m-6}{3+m^2}=\frac{7m-1}{m^2+3}\)
=>\(\frac{7m-1}{m^2+3}=1-\frac{m^2}{m^2+3}\Leftrightarrow\frac{7m-1}{m^2+3}=\frac{m^2+3}{m^2+3}-\frac{m^2}{m^2+3}\)
=>7m-1=m2+3-m2
=>7m-1=3
=>m=4/7
Vậm m=4/7