Ta có: \(\left\{{}\begin{matrix}2x+3y=m\\5x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=m\\15x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x=m+3\\5x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{17}\\y=5x-1=\dfrac{5m+15}{17}-\dfrac{17}{17}=\dfrac{5m-2}{17}\end{matrix}\right.\)

Để hệ phương trình có nghiệm duy nhất sao cho x<0 và y>0 thì 

\(\left\{{}\begin{matrix}\dfrac{m+3}{17}< 0\\\dfrac{5m-2}{17}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+3< 0\\5m-2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< -3\\m>\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow m\in\varnothing\)

\(2)mx^2-2\left(m-1\right)x+m-1=0\)

Để pt có nghiệm kép \(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\\Delta=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\\left[-2\left(m-1\right)\right]^2-4m\left(m-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow4\left(m^2-2m+1\right)-4m^2+4m=0\)

\(\Leftrightarrow4m^2-8m+4-4m^2+4m=0\)

\(\Leftrightarrow-4m+4=0\)

\(\Leftrightarrow m=1\)

Vậy để pt trên có nghiệm kép thì \(\left\{{}\begin{matrix}m\ne0\\m=1\end{matrix}\right.\)

bạn giải 1 giúp mình với

Ta co system: 2x - y = 5 va m x + 3y = 4. Tu phuong trinh thu nhat: y = 2x - 5. Thay vao phuong trinh thu hai: m x + 3(2x - 5) = 4, hay (m + 6)x = 19. Khi m \neq -6 thi x = 19/(m + 6), y = 2x - 5 = (8 - 5m)/(m + 6). Dieu kien x > y duoc giai nhu sau: - Neu m + 6 > 0 (m > -6) thi 19 > 8 - 5m, tu do m > -11/5. - Neu m + 6 < 0 (m < -6) thi 19 < 8 - 5m, tu do m < -6 (luon dung). Giao lai, de x > y thi m < -6 hoac m > -11/5. Luu y m = -6 lam he phuong trinh vo nghiem. Vay gia tri m thoa man la m < -6 hoac m > -11/5.

Vì \(\dfrac{2}{5}\ne\dfrac{1}{-3}\)

nên hệ có nghiệm duy nhất

\(\left\{{}\begin{matrix}2x+y=5\\5x-3y=-11m+29\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x+3y=15\\5x-3y=-11m+29\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}11x=15-11m+29=44-11m\\2x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-m+4\\y=5-2\left(-m+4\right)=5+2m-8=2m-3\end{matrix}\right.\)

Để x,y là độ dài hai cạnh góc vuông có cạnh huyền bằng \(\sqrt{10}\) thì \(x^2+y^2=10\)

=>\(\left(-m+4\right)^2+\left(2m-3\right)^2=10\)

=>\(m^2-8m+16+4m^2-12m+9=10\)

=>\(5m^2-20m+25-10=0\)

=>\(m^2-4m+3=0\)

=>(m-1)(m-3)=0

=>\(\left[{}\begin{matrix}m=1\\m=3\end{matrix}\right.\)

Vì \(\frac21<>\frac32\)

nên hệ luôn có nghiệm duy nhất

\(\begin{cases}2x+3y=m+1\\ x+2y=2m-8\end{cases}\Rightarrow\begin{cases}2x+3y=m+1\\ 2x+4y=4m-16\end{cases}\)

=>\(\begin{cases}2x+4y-2x-3y=4m-16-m-1\\ x+2y=2m-8\end{cases}\Rightarrow\begin{cases}y=3m-17\\ x=2m-8-2\left(3m-17\right)=2m-8-6m+34=-4m+26\end{cases}\)

x=3y

=>-4m+26=3(3m-17)

=>9m-51=-4m+26

=>13m=77

=>\(m=\frac{77}{13}\)

b: xy>0

=>(-4m+26)(3m-17)>0

=>(4m-26)(3m-17)<0

=>\(\frac{17}{3}

a, hệ\(\Leftrightarrow\)$\left \{ {{x>\frac{1}{2} } \atop {x<m+2}} \right.$

để hệ có nghiệm ⇒ m+2< $\frac{1}{2}$ ⇒ m<$\frac{-3}{2}$

a, Thay m=3 vào hpt ta có :

\(\left\{{}\begin{matrix}2x+3y=3\\-5x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=3\\-15x+3y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=3\\17x=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{17}\\y=\frac{43}{51}\end{matrix}\right.\)

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 )