Bài 2:

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

=>-4x-2y=3 và 8x+2y=-2

=>x=1/4; y=-2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)

=>y=6 và x-2=5/4

=>x=13/4; y=6

c: =>x+y=24 và 3x+y=78

=>-2x=-54 và x+y=24

=>x=27; y=-3

d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)

=>y+2=1 và x-1=25

=>x=26; y=-1

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/ Bạn tự giải

b/ Để hệ có nghiệm duy nhất thì:

\(\left(m-1\right)^2-1\ne0\Leftrightarrow\left[{}\begin{matrix}m\ne0\\m\ne2\end{matrix}\right.\)

Khi đó hệ tương đương: \(\left\{{}\begin{matrix}\left(m-1\right)x+y=3m-4\\\left(m-1\right)x+\left(m-1\right)^2y=m\left(m-1\right)\end{matrix}\right.\)

Trừ pt dưới cho trên: \(m\left(m-2\right)y=\left(m-2\right)^2\Rightarrow y=\frac{m-2}{m}\)

\(\Rightarrow x=m-\left(m-1\right)y=\frac{3m-2}{m}\)

\(x+y=3\Leftrightarrow\frac{3m-2}{m}+\frac{m-2}{m}=3\Leftrightarrow4m-4=3m\Rightarrow m=4\)

Hệ đã cho vô nghiệm khi

\(m+2=\dfrac{m+1}{3}\ne\dfrac{3}{4}\Leftrightarrow m=-\dfrac{5}{2}\)

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/ 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...

thay m=2 vào HPT ta có
\(\left\{{}\begin{matrix}x+2y=2+1\\2x+y=2.2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=3\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+4y=6\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y=2\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
vậy ..........
 

mọi người ơi giúp mình vs mai ktra r