mọi người ơi giúp mình vs mai ktra r

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{m}{3}\\-2y=\dfrac{10}{3}m+4\end{matrix}\right.\)

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

Để \(x^2+y^2=10\)

\(\Leftrightarrow\left(\dfrac{-m}{3}\right)^2+\left(\dfrac{-5x}{3}-2\right)^2=10\)

\(\Leftrightarrow\dfrac{m^2}{9}+\dfrac{25m^2}{9}+\dfrac{20m}{3}+4=10\)

\(\Leftrightarrow\dfrac{26m^2}{9}+\dfrac{20m}{3}-6=0\)

\(\Leftrightarrow\dfrac{26m^2}{9}+\dfrac{60m}{9}-\dfrac{54}{9}=0\)

\(\Leftrightarrow26m^2+60m-54=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-3\\m=\dfrac{9}{13}\end{matrix}\right.\)

a. Bạn tự giải

b. \(\left\{{}\begin{matrix}6x+2my=2m\\\left(m^2-m\right)x+2my=m^2-m\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6x+2my=2m\\\left(m^2-m-6\right)x=m^2-3m\end{matrix}\right.\)

Hệ có nghiệm duy nhất khi \(m^2-m-6\ne0\Rightarrow m\ne\left\{-2;3\right\}\)

Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{m}{m+2}\\y=\dfrac{m-1}{m+2}\end{matrix}\right.\) 

\(x+y^2=1\Leftrightarrow\dfrac{m}{m+2}+\left(\dfrac{m-1}{m+2}\right)^2=1\)

\(\Leftrightarrow m^2-4m-3=0\)

\(\Leftrightarrow...\)

a)

Khi m = 1, ta có:

{ x+2y=1+3   

  2x-3y=1

=> { x+2y=4

        2x-3y=1

=> { 2x+4y=8

        2x-3y=1

=> { x+2y=4

        2x-3y-2x-4y=1-8

=> { x=4-2y

       -7y = -7

=> { x = 2

        y = 1

Vậy khi m = 1 thì hệ phương trình có cặp nghệm

(x; y) = (2;1)

a) Thay m=1 vào HPT ta có: 

\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}2x+4y=8\\7y=7\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y)= (2;1)

a) Thay m=1 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=7\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4-2y=4-2=2\end{matrix}\right.\)

Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất là (x,y)=(2;1)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\2\left(m+3-2y\right)-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\2m+6-4y-3y-m=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\-7y+m+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\-7y=-m-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\y=\dfrac{m+6}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2\cdot\dfrac{m+6}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-\dfrac{2m+12}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7m+21-2m-12}{7}=\dfrac{5m+9}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\)

Để hệ phương trình có nghiệm duy nhất thỏa mãn x+y=3 thì \(\dfrac{5m+9}{7}+\dfrac{m+6}{7}=3\)

\(\Leftrightarrow6m+15=21\)

\(\Leftrightarrow6m=6\)

hay m=1

Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất thỏa mãn x+y=3

a/ Thay  \(m=1\) vào hpt ta có :

\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\)

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

Vậy...

b/ Ta có :

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

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

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

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

a/ Xét pt : \(\left\{{}\begin{matrix}mx-y=1\\\dfrac{x}{2}-\dfrac{y}{2}=335\end{matrix}\right.\)

 Khi \(m=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-y=670\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-669\\y=-1339\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}mx-y=1\\x-y=670\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-670\\mx-\left(x-670\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-670\\x\left(m-1\right)=-669\end{matrix}\right.\)

Để pt có nghiệm duy nhất \(\Leftrightarrow m\ne1\)

Vậy...