Cho số phức z thỏa mãn phương trình 4|z+i| + 3|z-i| = 10

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \mathrm{Ta} \mathrm{có}: \left|\mathrm{z}+\mathrm{i}\right|=\sqrt{{\mathrm{a}}^2+{\left(\mathrm{b}+1\right)}^2}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{b}+1}=\sqrt{{\left|\mathrm{z}\right|}^2+2\mathrm{b}+1} \\ \mathrm{Tương} \mathrm{tự}: \left|\mathrm{z}-\mathrm{i}\right|=\sqrt{{\left|\mathrm{z}\right|}^2-2\mathrm{b}+1} \\ \mathrm{Áp} \mathrm{dụng} \mathrm{BĐT} \mathrm{bunhiacopski} \mathrm{dạng} \left({\mathrm{a}}^2+{\mathrm{b}}^2\right)\left({\mathrm{c}}^2+{\mathrm{d}}^2\right)\le {\left(\mathrm{ac}+\mathrm{bd}\right)}^2 \\ \Rightarrow 4\left|\mathrm{z}+\mathrm{i}\right|+3\left|\mathrm{z}-\mathrm{i}\right|\le \sqrt{4^2+3^2}.\sqrt{{\left|\mathrm{z}+\mathrm{i}\right|}^2+{\left|\mathrm{z}-\mathrm{i}\right|}^2} \\ \iff 10\le 5\sqrt{{\left|\mathrm{z}\right|}^2+2\mathrm{b}+1+{\left|\mathrm{z}\right|}^2-2\mathrm{b}+1}=5\sqrt{2{\left|\mathrm{z}\right|}^2+2} \\ \iff 4\le 2{\left|\mathrm{z}\right|}^2+2 \\ \iff {\left|\mathrm{z}\right|}^2\ge 1\iff \left|\mathrm{z}\right|\ge 1 \\ \mathrm{KL}: {\left|\mathrm{z}\right|}_{\min }=1 \end{gathered}$$