Lời giải của Tự Học 365

Ta có \({{z}^{2}}-2z+5={{\left( z-1 \right)}^{2}}+4={{\left( z-1 \right)}^{2}}-{{\left( 2i \right)}^{2}}=\left( z-1+2i \right)\left( z-1-2i \right)\)

Do đó \(\left| {{z}^{2}}-2z+5 \right|=\left| \left( z-1+2i \right)\left( z+3i-1 \right) \right|\Leftrightarrow \left| \left( z-1+2i \right)\left( z-1-2i \right) \right|=\left| \left( z-1+2i \right)\left( z+3i-1 \right) \right|\)

\(\Leftrightarrow \left| z-1-2i \right|=\left| z+3i-1 \right|\Leftrightarrow \left| x-1+\left( y-2 \right)i \right|=\left| x-1+\left( y+3 \right)i \right|\) với \(z=x+yi\,\,\,\,\left( x,y\in \mathbb{R} \right).\)

\(\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( x-1 \right)}^{2}}+{{\left( y+3 \right)}^{2}}\Leftrightarrow y=-\dfrac{1}{2}\Rightarrow z=x-\dfrac{1}{2}i.\)

Khi đó \(\left| w \right|=\left| z-2+2i \right|=\left| x-2+\dfrac{3}{2}i \right|=\sqrt{{{\left( x-2 \right)}^{2}}+\dfrac{9}{4}}\ge \dfrac{3}{2}.\) Vậy \(\min \left| w \right|=\dfrac{3}{2}.\)

Đáp án cần chọn là: a