Cho hệ phương trình

- Hướng dẫn giải

$$\begin{gathered} \left\{\begin{array}{l}x+\left(m+1\right)y=1\\ 4x-y=-2\end{array}\right.\iff \left\{\begin{array}{l}y=4x+2\\ x+\left(m+1\right)\left(4x+2\right)=1\end{array}\right. \\ \iff \left\{\begin{array}{l}y=4x+2\\ x+4x\left(m+1\right)+2\left(m+1\right)=1\end{array}\right.\iff \left\{\begin{array}{l}y=4x+2\\ x\left(4m+5\right)=-\left(2m+1\right)\end{array}\right. \end{gathered}$$

Nếu $m=-\frac{5}{4}\Rightarrow 0x=\frac{3}{2}$(vô lý)

Nếu $m\ne -\frac{5}{4}\Rightarrow x=\frac{-2m-2}{4m+5}\Rightarrow y=4x+2=\frac{6}{4m+5}$

Theo bài ra: $x^2+y^2=\frac{1}{4}\Rightarrow {\left(\frac{-2m-1}{4m+5}\right)}^2+{\left(\frac{6}{4m+5}\right)}^2=\frac{1}{4}$

$$\begin{gathered} \iff 4(4m^2+4m+1+36)=16m^2+40m+25 \\ \iff 24m=124\iff m=\frac{41}{8} \end{gathered}$$

Đáp án:A