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Câu hỏi: Cho số phức z thỏa mãn $\left( 2-i \right)\left| z \right|-\dfrac{25}{\overline{z}}=6-2i.$ Khi đó $\left| z \right|$ thuộc khoảng nào trong các khoảng sau?
A. $\left( 2;4 \right).$
B. $\left( 4;6 \right).$
C. $\left( 9;11 \right).$
D. $\left( 11;14 \right).$
A. $\left( 2;4 \right).$
B. $\left( 4;6 \right).$
C. $\left( 9;11 \right).$
D. $\left( 11;14 \right).$
Điều kiện bài toán tương đương:
$\left( 2-i \right)\left| z \right|-6+2i=\dfrac{25}{\overline{z}}\Leftrightarrow \left( 2\left| z \right|-6 \right)-\left( \left| z \right|-2 \right)i=\dfrac{25}{\overline{z}}.$
$\Rightarrow \left( 2\left| z \right|-6 \right)-\left( \left| z \right|-2 \right)i=\left| \dfrac{25}{\overline{z}} \right|\Leftrightarrow \sqrt{{{\left( 2\left| z \right|-6 \right)}^{2}}+{{\left( \left| z \right|-2 \right)}^{2}}}=\dfrac{25}{\left| \overline{z} \right|}$ (*)
Đặt $t=\left| z \right|>0,$ khi đó (*) có dạng: $\sqrt{{{\left( 2t-6 \right)}^{2}}+{{\left( t-2 \right)}^{2}}}=\dfrac{25}{t}\Leftrightarrow 5{{t}^{2}}-28t+40=\dfrac{625}{{{t}^{2}}}$
$\Leftrightarrow 5{{t}^{4}}-28{{t}^{3}}+40{{t}^{2}}-625=0\Leftrightarrow \left( t-5 \right)\left( 5{{t}^{3}}-3{{t}^{2}}+25t+125 \right)=0$ (2*)
Do $\left( 5{{t}^{3}}-3{{t}^{2}}+25t+125 \right)=0\Leftrightarrow t\left( 5{{t}^{2}}-3t+25 \right)+125>0,\forall t>0,$ suy ra:
(2*) $\Leftrightarrow t=5\Leftrightarrow \left| z \right|=5\in \left( 4;6 \right).$
$\left( 2-i \right)\left| z \right|-6+2i=\dfrac{25}{\overline{z}}\Leftrightarrow \left( 2\left| z \right|-6 \right)-\left( \left| z \right|-2 \right)i=\dfrac{25}{\overline{z}}.$
$\Rightarrow \left( 2\left| z \right|-6 \right)-\left( \left| z \right|-2 \right)i=\left| \dfrac{25}{\overline{z}} \right|\Leftrightarrow \sqrt{{{\left( 2\left| z \right|-6 \right)}^{2}}+{{\left( \left| z \right|-2 \right)}^{2}}}=\dfrac{25}{\left| \overline{z} \right|}$ (*)
Đặt $t=\left| z \right|>0,$ khi đó (*) có dạng: $\sqrt{{{\left( 2t-6 \right)}^{2}}+{{\left( t-2 \right)}^{2}}}=\dfrac{25}{t}\Leftrightarrow 5{{t}^{2}}-28t+40=\dfrac{625}{{{t}^{2}}}$
$\Leftrightarrow 5{{t}^{4}}-28{{t}^{3}}+40{{t}^{2}}-625=0\Leftrightarrow \left( t-5 \right)\left( 5{{t}^{3}}-3{{t}^{2}}+25t+125 \right)=0$ (2*)
Do $\left( 5{{t}^{3}}-3{{t}^{2}}+25t+125 \right)=0\Leftrightarrow t\left( 5{{t}^{2}}-3t+25 \right)+125>0,\forall t>0,$ suy ra:
(2*) $\Leftrightarrow t=5\Leftrightarrow \left| z \right|=5\in \left( 4;6 \right).$
Đáp án B.