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Câu hỏi: Xét số phức z thỏa mãn $\left( 1+2i \right)\left| z \right|=\dfrac{\sqrt{10}}{z}-2+i$. Mệnh đề nào dưới đây đúng?
A. $\dfrac{1}{2}<\left| z \right|<\dfrac{3}{2}$
B. $\dfrac{3}{2}<\left| z \right|<2$
C. $\left| z \right|>2$
D. $\left| z \right|>\dfrac{1}{2}$
A. $\dfrac{1}{2}<\left| z \right|<\dfrac{3}{2}$
B. $\dfrac{3}{2}<\left| z \right|<2$
C. $\left| z \right|>2$
D. $\left| z \right|>\dfrac{1}{2}$
$\left( 1+2i \right)\left| z \right|=\dfrac{\sqrt{10}}{z}-2+i\Leftrightarrow \left| z \right|+2+\left( 2\left| z \right|-1 \right)i=\dfrac{\sqrt{10}}{z}$
$\Rightarrow \left| \left| z \right|+2+\left( 2\left| z \right|-1 \right)i \right|=\left| \dfrac{\sqrt{10}}{z} \right|\Leftrightarrow \sqrt{{{\left( \left| z \right|+2 \right)}^{2}}+{{\left( 2\left| z \right|-1 \right)}^{2}}}=\dfrac{\sqrt{10}}{\left| z \right|}$
$\Leftrightarrow {{\left( \left| z \right|+2 \right)}^{2}}+{{\left( 2\left| z \right|-1 \right)}^{2}}=\dfrac{10}{{{\left| z \right|}^{2}}}\Leftrightarrow 5{{\left| z \right|}^{4}}+5{{\left| z \right|}^{2}}-10=0\Rightarrow \left| z \right|=1$
Vậy $\dfrac{1}{2}<\left| z \right|<\dfrac{3}{2}$
$\Rightarrow \left| \left| z \right|+2+\left( 2\left| z \right|-1 \right)i \right|=\left| \dfrac{\sqrt{10}}{z} \right|\Leftrightarrow \sqrt{{{\left( \left| z \right|+2 \right)}^{2}}+{{\left( 2\left| z \right|-1 \right)}^{2}}}=\dfrac{\sqrt{10}}{\left| z \right|}$
$\Leftrightarrow {{\left( \left| z \right|+2 \right)}^{2}}+{{\left( 2\left| z \right|-1 \right)}^{2}}=\dfrac{10}{{{\left| z \right|}^{2}}}\Leftrightarrow 5{{\left| z \right|}^{4}}+5{{\left| z \right|}^{2}}-10=0\Rightarrow \left| z \right|=1$
Vậy $\dfrac{1}{2}<\left| z \right|<\dfrac{3}{2}$
Đáp án A.