Thay x = 1; y = 3 vào hệ ta có

$$\begin{gathered} \left\{\begin{array}{l}2.1+b.3=a\\ b.1+a.3=5\end{array}\right.\iff \left\{\begin{array}{l}a-3b=2\\ 3a+b=5\end{array}\right.\iff \left\{\begin{array}{l}3a-9b=6\\ 3a+b=5\end{array}\right. \\ \iff \left\{\begin{array}{l}10b=-1\\ 3a+b=5\end{array}\right.\iff \left\{\begin{array}{l}b=\frac{-1}{10}\\ a=\frac{17}{10}\end{array}\right. \end{gathered}$$

Vậy $a=\frac{17}{10};b=\frac{-1}{10}$thì hệ phương trình có nghiệm  x = 1; y = 3$\Rightarrow$10(a + b) = 16

Đáp án: B

Thay x = −1; y = −2 vào hệ ta có:

$$\begin{gathered} \left\{\begin{array}{l}3a\left(-1\right)+\left(-2\right)=b\\ 2.a\left(-1\right)-2b\left(-2\right)=3\end{array}\right.\iff \left\{\begin{array}{l}-3a-2=b\\ -2a+4b=3\end{array}\right. \\ \iff \left\{\begin{array}{l}b=-2-3a\\ -2a+4\left(-2-3a\right)=3\end{array}\right.\iff \left\{\begin{array}{l}b=-2-3a\\ 14a=-11\end{array}\right. \\ \iff \left\{\begin{array}{l}a=-\frac{11}{14}\\ b=-2-3.\left(-\frac{11}{14}\right)\end{array}\right.\iff \left\{\begin{array}{l}a=-\frac{11}{14}\\ b=\frac{5}{14}\end{array}\right. \end{gathered}$$

Vậy $a=\frac{-11}{14};b=\frac{5}{14}$thì hệ phương trình có nghiệm  x = −1; y = −2

$\Rightarrow$14(a – b) = $14.\left(-\frac{11}{14}-\frac{5}{14}\right)$ = −16

Đáp án: C

Ta có

$$\begin{gathered} \left\{\begin{array}{l}x+2y=m+3\\ 2x-3y=m\end{array}\right.\iff \left\{\begin{array}{l}2x+4y=2m+6\\ 2x-3y=m\end{array}\right. \\ \iff \left\{\begin{array}{l}x+2y=m+3\\ 7y=m+6\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{5m+9}{7}\\ y=\frac{m+6}{7}\end{array}\right. \end{gathered}$$

Hệ phương trình có nghiệm duy nhất $(x; y) =\left(\frac{5m+9}{7};\frac{m+6}{7}\right)$  

Lại có x + y = −3 hay $\frac{5m+9}{7}+\frac{m+6}{7}=-3\iff$5m + 9 + m + 6 = −21

$\iff$6m = −36$\iff$m = −6

Vậy với m = −6 thì hệ phương trình có nghiệm duy nhất (x; y) thỏa mãn x + y = −3

Đáp án: A

Ta có

$$\begin{gathered} \left\{\begin{array}{l}3x-y=2m+1\\ x+2y=-m+2\end{array}\right.\iff \left\{\begin{array}{l}6x-2y=4m+2\\ x+2y=-m+2\end{array}\right. \\ \iff \left\{\begin{array}{l}7x=3m+4\\ x+2y=-m+2\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{3m+4}{7}\\ \frac{3m+4}{7}+2y=-m+2\end{array}\right. \\ \iff \left\{\begin{array}{l}x=\frac{3m+4}{7}\\ 2y=\frac{-7m+14}{7}-\frac{3m+4}{7}\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{3m+4}{7}\\ y=\frac{-5m+5}{7}\end{array}\right. \end{gathered}$$

hệ phương trình có nghiệm duy nhất $(x; y) =\left(\frac{3m+4}{7};\frac{-5m+5}{7}\right)$  

Để x – y = 1 thì$\frac{3m+4}{7}-\frac{-5m+5}{7}=1\iff$8m – 1 = 7$\iff$8m = 8 $\iff$  m = 1

Vậy với m = 1 thì hệ phương trình có nghiệm duy nhất (x; y) thỏa mãn x − y = 1

Đáp án: C

Ta có $\left\{\begin{array}{l}2x+y=5m-1\\ x-2y=2\end{array}\right.$

$\iff \left\{\begin{array}{l}y=5m-1-2x\\ x-2\left(5m-1-2x\right)=2\end{array}\right.\iff \left\{\begin{array}{l}y=5m-1-2x\\ 5x=10m\end{array}\right.$

$\iff \left\{\begin{array}{l}x=2m\\ y=m-1\end{array}\right.$

Thay vào $x^2 – 2y^2 = -2$ ta có

$$\begin{gathered} x^2–2y^2=-2\iff {\left(2m\right)}^2–2{(m-1)}^2 =-2 \\ \iff 2m^2+4m=0\iff \left[\begin{array}{l}m=0\\ m=-2\end{array}\right. \end{gathered}$$    

Vậy m $\in${−2; 0}

Đáp án: C