a) Khi \(k=1\) ta có hệ phương trình: \(\left\{{}\begin{matrix}x+y=1\\2x-y=5\end{matrix}\right.\)

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

Vậy hệ có nghiệm \(\left(x;y\right)=\left(2;-1\right)\).

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

Điều kiện: \(y+1\ne0\Leftrightarrow y\ne-1\Leftrightarrow2k-3\ne-1\Leftrightarrow k\ne1\)

\(\dfrac{x^2-y-5}{y+1}=4\Leftrightarrow x^2-y-5=4y+4\\ \Leftrightarrow\left(k+1\right)^2-\left(2k-3\right)-5=4\left(2k-3\right)+4\\ \Leftrightarrow k^2+2k+1-2k+3-5=8k-12+4\\ \Leftrightarrow k^2-8k+7=0\Leftrightarrow\left(k-1\right)\left(k-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}k-1=0\\k-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k=1\\k=7\end{matrix}\right.\)

Kết hợp điều kiện \(k\ne1\) ta được \(k=7\) là giá trị cần tìm.

a)Khi k = 1 thì ta có hệ phương trình:

    \(\left\{{}\begin{matrix}x+y=3.1-2\\2x-y=5\end{matrix}\right.\)

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

⇔\(\left\{{}\begin{matrix}3x=6\\x+y=1\end{matrix}\right.\)

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

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

Vậy ...

Câu 10: B

Câu 9: C

Câu 8: A

Câu 7: A

Câu 6: C

Câu 5:D

Câu 4: A

Câu 3: B

Câu 2: A

Câu 1; B

\(\left\{{}\begin{matrix}2x+ky=1\\kx+2y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\k\left(-\dfrac{k}{2}y+\dfrac{1}{2}\right)+2y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\end{matrix}\right.\)

Hệ PT có nghiệm \(\Leftrightarrow\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\) có nghiệm

\(\Leftrightarrow-\dfrac{k^2}{2}+2\ne0\Leftrightarrow\dfrac{k^2}{2}=2\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)

\(m_1=\dfrac{11}{2}\)

\(m_2=3\)

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 

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

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