Câu hỏi 30

Đặt \(z=a+bi\Rightarrow \overline{z}=a-bi\,\,\left( a;b\in \mathbb{R} \right)\).

\(\Rightarrow P=\sqrt{{{\left( a-2 \right)}^{2}}+{{\left( b+2 \right)}^{2}}}+\sqrt{{{\left( a+2 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}+\sqrt{{{\left( a+3 \right)}^{2}}+{{\left( b+3 \right)}^{2}}}=f\left( a;b \right)\).

Ta có \(f\left( a;b \right)=f\left( b;a \right)\,\,\forall a,b\), ta dự đoán dấu "=" xảy ra \(\Leftrightarrow a=b=k\).

\(\sqrt{{{\left( a-2 \right)}^{2}}+{{\left( b+2 \right)}^{2}}}\ge \frac{1}{\sqrt{{{m}^{2}}+{{n}^{2}}}}\sqrt{\left( {{m}^{2}}+{{n}^{2}} \right)\left[ {{\left( a-2 \right)}^{2}}+{{\left( b+2 \right)}^{2}} \right]}\ge \frac{m\left( a-2 \right)+n\left( b+2 \right)}{\sqrt{{{m}^{2}}+{{n}^{2}}}}\).

Dấu "=" xảy ra \(\Leftrightarrow \left\{ \begin{align}\frac{m}{a-2}=\frac{n}{b+2} \\ a=b=k \\ \end{align} \right.\). Chọn \(m=k-2,\,\,n=k+2\).

\(\Rightarrow \sqrt{{{\left( a-2 \right)}^{2}}+{{\left( b+2 \right)}^{2}}}\ge \frac{\left( k-2 \right)\left( a-2 \right)+\left( k+2 \right)\left( b+2 \right)}{\sqrt{2{{k}^{2}}+8}}=\frac{k\left( a+b \right)-2a+2b+8}{\sqrt{2{{k}^{2}}+8}}\)

Tương tự :

\(\begin{align}\sqrt{{{\left( a+2 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}\ge \frac{k\left( a+b \right)+2a-2b+8}{\sqrt{2{{k}^{2}}+8}} \\ \sqrt{{{\left( a+3 \right)}^{2}}+{{\left( b+3 \right)}^{2}}}\ge \frac{1}{\sqrt{2}}\sqrt{{{\left[ 1\left( a+3 \right)+1\left( b+3 \right) \right]}^{2}}}=\frac{a+b+6}{\sqrt{2}} \\ \end{align}\)

Cộng vế với vế ta có: \(P\ge \frac{2k\left( a+b \right)}{\sqrt{2{{k}^{2}}+8}}+\frac{16}{\sqrt{2{{k}^{2}}+8}}+\frac{a+b}{\sqrt{2}}+\frac{6}{\sqrt{2}}\) cần chọn số \(k\) sao cho \(\frac{2k}{\sqrt{2{{k}^{2}}+8}}+\frac{1}{\sqrt{2}}=0\Leftrightarrow k=-\frac{2}{\sqrt{3}}\).

Khi đó \(P\ge 2\sqrt{6}+3\sqrt{2}\).

Vậy \(m=q=2;\,\,n=6;\,\,p=3 \,\,\Rightarrow m+n+p+q=2+6+3+2=13\).

Chọn B.

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