Bài toán này khó lắm , ai giải hộ cái

C

conga222222

$\eqalign{
& z = a + bi \cr
& {\left| {2z + 1} \right|^2} = {\left( {2a + 1} \right)^2} + {\left( {2b} \right)^2} \cr
& {\left| {z + \overline z + 3} \right|^2} = {\left( {2a + 3} \right)^2} \cr
& \left| {2x + 1} \right| = \left| {z + \overline z + 3} \right| \leftrightarrow {\left| {2z + 1} \right|^2} = {\left| {z + \overline z + 3} \right|^2} \leftrightarrow {\left( {2a + 1} \right)^2} + {\left( {2b} \right)^2} = {\left( {2a + 3} \right)^2} \cr
& \leftrightarrow 4{b^2} = 8a + 8 \cr
& \leftrightarrow {b^2} = a + 1 \cr
& co\;{\left| w \right|^2} = {\left( {a - 8} \right)^2} + {b^2} = {a^2} - 15a + 65 = {\left( {a - {{15} \over 2}} \right)^2} + {{35} \over 4} \ge {{35} \over 4} \cr
& \to \left| w \right| \ge {{\sqrt {35} } \over 2} \cr
& dau = \leftrightarrow a = {{15} \over 2}\;b = \pm \sqrt {{{17} \over 2}} \cr} $
 
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