Chọn A
Gọi \(z-a+b.i\, \, (a,\, b\in {\rm R})\)
Theo đề bài
\(\left|2z+2-3i\right|=1\Leftrightarrow \left|2a+2bi+2-3i\right|=1\)
\(\Leftrightarrow \left|2a+2+\left(2b-3\right)i\right|=1\)
\(
\Leftrightarrow \left(2a+2\right)^{2} +\left(2b-3\right)^{2} =1\)
\(\Leftrightarrow \left(a+1\right)^{2} +\left(b-\frac{3}{2} \right)^{2} =\frac{1}{4} (1)
\)
\(\Leftrightarrow a^{2} +2a+1+b^{2} -3b+\frac{9}{4} =\frac{1}{4} \)
\(\Leftrightarrow a^{2} +b^{2} =-2a+3b-3.\)
Từ \((1)\) suy ra
\(\left(b-\frac{3}{2} \right)^{2} \le \frac{1}{4} \Rightarrow -\frac{1}{2} \le b-\frac{3}{2} \le \frac{1}{2} \Rightarrow 1\le b\le 2.\)
Ta có \(2\left|z+2\right|+\left|z-3\right|=2\left|a+2+bi\right|+\left|a-3+bi\right|\)
\(=2\sqrt{\left(a+2\right)^{2} +b^{2} } +\sqrt{\left(a-3\right)^{2} +b^{2} } \)
\(=2\sqrt{a^{2} +4a+4+b^{2} } +\sqrt{a^{2} -6a+9+b^{2} } \)
\(=2\sqrt{2a+3b+1} +\sqrt{-8a+3b+6} \)
Áp dụng bất đẳng thức Bunhiaxcopki cho 1
\(,\, \sqrt{4\left(2a+3b+1\right)} ,\, 1,\, \sqrt{-8a+3b+6}\) ta có
\(2\sqrt{2a+3b+1} +\sqrt{-8a+3b+6} \)
\(\le \sqrt{2.\left(8a+12b+4-8a+3b+6\right)}\)
\( \le \sqrt{2\left(15b+10\right)} \le 4\sqrt{5} \)
Dấu bằng xảy ra khi
\(\left\{\begin{array}{l} {b=2} \\ {\frac{1}{\sqrt{8a+12b+4} } =\frac{1}{\sqrt{-8a+3b+6} } } \end{array}\right. \left\{\begin{array}{l} {b=2} \\ {8a+12b+4=-8a+3b+6} \end{array}\right. \)
\(\Leftrightarrow \left\{\begin{array}{l} {b=2} \\ {a=-1} \end{array}\right. .\)
Do đó a-b=-3.
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