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

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

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b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\y=2m-1-3x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=-m-1\end{matrix}\right.\)

Ta có: \(x^2+y^2=5\) 

\(\Rightarrow m^2+m^2+2m+1=5\) \(\Leftrightarrow m^2+m-2=0\) \(\Rightarrow\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)

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c) Hệ phương trình luôn có nghiệm duy nhất

Ta có: \(x-3y>0\)

\(\Rightarrow m-3\left(-m-1\right)>0\)

\(\Leftrightarrow4m+3>0\) \(\Leftrightarrow m>-\dfrac{3}{4}\)

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a) Thay m=1 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}3x-y=1\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=1\\3x+6y=15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-7y=-14\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=5-2y=5-2\cdot2=1\end{matrix}\right.\)

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

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 )

a) Thay m=1 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}3x-y=1\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=1\\3x+6y=15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-7y=-14\\3x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

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

a:

Để hệ có nghiệm duy nhất thì m/2<>-2/-m

=>m^2<>4

=>m<>2 và m<>-2

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

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

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

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

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

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

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

Để \(x^2+y^2=10\)

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

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

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

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

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

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