Ta có $z=a+bi(a,b\in ℝ)\Rightarrow \overline{z}=a-bi$

Vậy $(2+i)(\overline{z}+1-i)-(2-3i)(z+i)=2+5i\iff (2+i)(a-bi+1-i)-(2-3i)(a+bi+i)=2+5i$

$$\begin{gathered} \iff (2+i)(a+1-(b+1\left)i\right)-(2-3i)(a+(b+1\left)i\right)=2+5i \\ \iff 2(\text{a}+1)-2( \text{b}+1)\text{i}+(\text{a}+1)\text{i}-(\text{b}+1){\text{i}}^2-\left(2\text{a}+2( \text{b}+1)\text{i}-3\text{a}-3( \text{b}+1){\text{i}}^2\right)=2+5\text{i} \\ \iff 2a+2-(2b+2)i+(a+1)i+b+1-2a-(2b+2)i+3ai-3b-3=2+5i \\ \iff -2b+(4a-4b-3)i=2+5i\iff \left\{\begin{array}{l}-2b=2\\ 4a-4b-3=5\end{array}\right. \\ \iff \left\{\begin{array}{l}a=1\\ b=-1\end{array}\right. \end{gathered}$$