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 )

1)

\(\left\{{}\begin{matrix}x+y=4\\2x+3y=m\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+3y=12\\2x+3y=m\end{matrix}\right.\)

trừ 2 vế của pt cho nhau ta tìm được

\(\left\{{}\begin{matrix}x=12-m\\y=m-8\end{matrix}\right.\)

để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m< 12\\m< 8\end{matrix}\right.\Rightarrow}m< 8}\)

\(a) \begin{cases}x=y+4\\2x+3=0\end{cases}\Leftrightarrow\begin{cases}x = y + 4\\2x = -3\end{cases}\Leftrightarrow\begin{cases}\dfrac{-3}{2} = y + 4\\x = \dfrac{-3}{2}\end{cases}\Leftrightarrow\begin{cases}y = \dfrac{-11}{2}\\x = \dfrac{-3}{2}\end{cases}\\b) \begin{cases}2x + y = 7\\3y - x = 7\end{cases}\Leftrightarrow\begin{cases}2x + y = 7\\6y - 2x = 14\end{cases}\Leftrightarrow\begin{cases}2x + y = 7\\7y = 21\end{cases}\Leftrightarrow\begin{cases}2x + 3 = 7\\y = 3\end{cases}\Leftrightarrow\begin{cases}x=2\\y=3\end{cases}\\ c) \begin{cases} 5x + y = 3 \\ -x - \dfrac{1}{5}y=\dfrac{-3}{5} \end{cases} \Leftrightarrow \begin{cases} 5x + y = 3 \\ 5x + y = 3 \end{cases} (luôn\ đúng) \Leftrightarrow Phương\ trình\ vô\ số\ nghiệm \\d) \begin{cases} 3x - 5y = -18 \\ x - 5 = 2y \end{cases} \Leftrightarrow \begin{cases} 3x - 5y = -18 \\ 3x - 6y = 15 \end{cases} \Leftrightarrow \begin{cases} x - 5 = 2.(-33)\\ y = -13 \end{cases} \Leftrightarrow \begin{cases}x = -61\\y=-33 \end{cases} \)

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne-\dfrac{1}{m}\)

=>\(-m^2\ne2\)(luôn đúng)

\(\left\{{}\begin{matrix}mx-y=m^2\\2x+my=m^2+2m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-m^2\\2x+m\left(mx-m^2\right)=m^2+2m+2\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}y=mx-m^2\\x\left(m^2+2\right)=\left(m+1\right)\left(m^2+2\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=mx-m^2=m\left(m+1\right)-m^2=m\end{matrix}\right.\)

Đặt \(A=x^2+3y+4\)

\(=\left(m+1\right)^2+3m+4\)

\(=m^2+5m+5\)

\(=m^2+2\cdot m\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}\)

\(=\left(m+\dfrac{5}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu '=' xảy ra khi m=-5/2

Câu 1: 

Lấy PT(1) + PT(2) theo vế thu được:

$3x+y+(2x-y)=10$

$\Leftrightarrow 5x=10$

$\Leftrightarrow x=2$

$y=2x-7=2.2-7=-3$

Vậy hpt có nghiệm $(x,y)=(2,-3)$

Câu 2: 

Lấy PT(1) - PT(2) theo vế thì:

$(2x+5y)-(2x-3y)=8$

$\Leftrightarrow 8y=8$

$\Leftrightarrow y=1$

Khi đó: $x=3y:2=\frac{3}{2}$
Vậy.............

Câu 3:

Lấy PT(1) - 2PT(2) thu được:

$(4x+3y)-2(2x+y)=6-2.4$

$\Leftrightarrow y=-2$

Khi đó:

$2x=4-y=6$

$\Leftrightarrow x=3$

Vậy..........

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

\(\Rightarrow17x=m+3\)

\(\Leftrightarrow x=\dfrac{m+3}{17}\)

để x>0 \(\Leftrightarrow\dfrac{m+3}{17}>0\Leftrightarrow m+3>0\Leftrightarrow m>-3\)

còn y> gì bạn cũng làm như zậy nhé :))