Bài 3         $\left\{\begin{array}{l}(m-1)x-my=3m-1 \left(1\right)\\ 2x-y=m+5 \left(2\right)\end{array}\right.$
a)
           Thay m=2 vào hệ phương trình ta có:
$\left\{\begin{array}{l}\mathrm{x}-2\mathrm{y}=5\\ 2\mathrm{x}-\mathrm{y}=7\end{array}\right.$
$\leftrightarrow \left\{\begin{array}{l}\mathrm{x}=3\\ \mathrm{y}=-1\end{array}\right.$
  Vậy hệ phương trình có nghiệm duy nhất $\left\{\begin{array}{l}\mathrm{x}=3\\ \mathrm{y}=-1\end{array}\right.với m=2$
b)
           Từ $\left(2\right)$ ta có:
                           $\mathrm{y}=2\mathrm{x}-\mathrm{m}-5$
             Thay $\mathrm{y}=2\mathrm{x}-\mathrm{m}-5$​ vào $\left(1\right)$ ta có:
$$\begin{gathered} (\mathrm{m}-1)\mathrm{x}-\mathrm{m}(2\mathrm{x}-\mathrm{m}-5)=3\mathrm{m}-1 \\ \leftrightarrow \mathrm{mx}-\mathrm{x}-2\mathrm{mx}+{\mathrm{m}}^2+5\mathrm{m}-3\mathrm{m}+1=0 \\ \leftrightarrow -\mathrm{mx}-\mathrm{x}+{\mathrm{m}}^2+2\mathrm{m}+1=0 \\ \leftrightarrow \mathrm{x}(\mathrm{m}+1)={(\mathrm{m}+1)}^2 (*) \end{gathered}$$
               Hệ phương trình có nghiệm duy nhất
                          ​  $\leftrightarrow$phương trình (*) có nghiệm duy nhất
                              $\leftrightarrow$ $\mathrm{m}+1 \mathrm{khác} 0$  
                                 $\leftrightarrow \mathrm{m} \mathrm{khác} -1$
$$\begin{gathered} \Rightarrow \mathrm{x}=\frac{{(\mathrm{m}+1)}^2}{\mathrm{m}+1}= \mathrm{m}+1 \\ \Rightarrow \mathrm{y}=\mathrm{m}-3 \\ \Rightarrow \mathrm{hệ} \mathrm{phương} \mathrm{trình} \mathrm{có} \mathrm{nghiệm} \mathrm{duy} \mathrm{nhất}\left\{\begin{array}{l}\mathrm{x}=\mathrm{m}+1\\ \mathrm{y}=\mathrm{m}-3\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}>0\\ \mathrm{y}<0\end{array}\right. \\ \iff \left\{\begin{array}{l}\mathrm{m}+1>0\\ \mathrm{m}-3<0\end{array}\right. \\ \iff \left\{\begin{array}{l}\mathrm{m}>-1\\ \mathrm{m}<3\end{array}\right. \\ \iff -1<\mathrm{m}<3 \\ \mathrm{Mà} \mathrm{m} \in \mathrm{Z} \\ \Rightarrow \mathrm{m}\in \{ 0;1;2\} \\ \mathrm{Vậy}... \end{gathered}$$
c)
           Theo câu b ta có   hệ phương trình có nghiệm duy nhất ...
$$\begin{gathered} \mathrm{P}={\mathrm{x}}^2+{\mathrm{y}}^2 \\ ( \mathrm{Thay} \mathrm{x} \mathrm{và} \mathrm{y} \mathrm{vào} \mathrm{P} \mathrm{sau} \mathrm{đó} \mathrm{sẽ} \mathrm{đc} \mathrm{kết} \mathrm{quả} \mathrm{là}:) \\ ={(\mathrm{m}-1)}^2+4 \mathrm{ta} \mathrm{có} {(\mathrm{m}-1)}^{2 }>=0 \Rightarrow {(\mathrm{m}-1)}^{2 }+4>=4 \\ \Rightarrow \mathrm{P}>=4 \\ \Rightarrow \mathrm{Min} \mathrm{P}=4\iff (\mathrm{m}-1)^2=0 \\ \iff \mathrm{m}-1=0 \\ \iff \mathrm{m}=1 \\ \mathrm{Vậy}.... \end{gathered}$$
Bài 4          
                    $$\begin{gathered} \mathrm{Gọi} \mathrm{độ} \mathrm{dài} 2 \mathrm{cạnh} \mathrm{của} \mathrm{tam} \mathrm{giác} \mathrm{vuông} \mathrm{đó} \mathrm{lần} \mathrm{lượt} \mathrm{là} \mathrm{a} \mathrm{và} \mathrm{b}\left(\mathrm{cm}\right)(\mathrm{ĐK}: \mathrm{a};\mathrm{b}>0) \\ \mathrm{Khi} \mathrm{đó} \mathrm{diện} \mathrm{tích} \mathrm{của} \mathrm{tam} \mathrm{giác} \mathrm{vuông} \mathrm{đó} \mathrm{là} 1/2\mathrm{ab} \left(\mathrm{cm}2\right) \\ \mathrm{Nếu} \mathrm{tăng} 2 \mathrm{cạnh} \mathrm{góc} \mathrm{vuông} \mathrm{lên} 4\mathrm{cm} \mathrm{và} 5\mathrm{cm} \mathrm{khi} \mathrm{đó} \mathrm{ta} \mathrm{đc} \mathrm{diện} \mathrm{tích} \mathrm{là} 1/2(\mathrm{a}+4)(\mathrm{b}+5) \mathrm{thì} \mathrm{diện} \mathrm{tích} \mathrm{tăng} \mathrm{thêm} 110\mathrm{cm}2 \mathrm{nên} \mathrm{ta} \mathrm{có} \mathrm{pt}: \\ 1/2(\mathrm{a}+4)(\mathrm{b}+5)-1/2\mathrm{ab}=110 \\ \iff 5\mathrm{a}+4\mathrm{b}=200 \left(1\right) \\ \mathrm{Nếu} \mathrm{giảm} 2 \mathrm{cạnh} \mathrm{này} \mathrm{đi} 5\mathrm{cm} \mathrm{khi} \mathrm{đó} \mathrm{ta} \mathrm{đc} \mathrm{điện} \mathrm{tích} \mathrm{là} 1/2(\mathrm{a}-5)(\mathrm{b}-5) \mathrm{thì} \mathrm{diện} \mathrm{tích} \mathrm{giảm} \mathrm{đi} 100\mathrm{cm}2 \mathrm{nên} \mathrm{ta} \mathrm{có} \mathrm{pt}: \\ 1/2\mathrm{ab}-1/2(\mathrm{a}-5)(\mathrm{b}-5)=100 \\ \iff \mathrm{a}+\mathrm{b}=45 \left(2\right) \\ \mathrm{Từ} 1 \mathrm{và} 2 \mathrm{ta} \mathrm{có} \mathrm{hệ} \mathrm{pt}: \\ \left\{\begin{array}{l}5\mathrm{a}+4\mathrm{b}=200\\ \mathrm{a}+\mathrm{b}=45\end{array}\right. \end{gathered}$$      ​                  (ra kết quả bằng 20 vs 25 nhé)  
                       ( Hơi tắt tí mog bạn thông cảm nhé !!!!)