đặc \(z=a+bi\) với \(a;b\in R;i^2=-1\)
ta có : \(\left|z-4-3i\right|=\sqrt{5}\Leftrightarrow\left(a-4\right)^2+\left(b-3\right)^2=5\)
\(\Leftrightarrow a^2+b^2=8x+6x-20\)
đặc \(A=\left|z+1-3i\right|+\left|z-1+i\right|=\sqrt{\left(a+1\right)^2+\left(b-3\right)^2}+\sqrt{\left(a-1\right)^2+\left(b+1\right)^2}\)
áp dụng bunhiacopxki ta có :
\(A\le\sqrt{2\left[\left(a+1\right)^2+\left(b-3\right)^2+\left(a-1\right)^2+\left(b+1\right)^2\right]}\)
\(\Leftrightarrow A\le\sqrt{2\left(2a^2+2b^2-4b+12\right)}=\sqrt{2\left(16a+12b-40-4b+12\right)}\)
\(\Leftrightarrow A\le\sqrt{2\left[16\left(a-4\right)+8\left(b-3\right)\right]+120}\)
áp dụng bunhiacopxki lần nữa ta có :
\(A\le\sqrt{2\left(16^2+8^2\right)\left[\left(a-4\right)^2+\left(b-3\right)^2\right]+120}\)
\(\Leftrightarrow A\le2\sqrt{830}\) dâu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(a+1\right)^2+\left(b-3\right)^2=\left(a-1\right)^2+\left(b+1\right)^2\\\dfrac{a-4}{16}=\dfrac{b-3}{8}\\\left(a-4\right)^2+\left(b-3\right)^2=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=6\\b=4\end{matrix}\right.\\\left\{{}\begin{matrix}a=2\\b=2\end{matrix}\right.\end{matrix}\right.\)
khi \(\left\{{}\begin{matrix}a=6\\b=4\end{matrix}\right.\Rightarrow P=a+b=10\)
khi \(\left\{{}\begin{matrix}a=2\\b=2\end{matrix}\right.\Rightarrow P=a+b=4\)
vậy \(P=10;P=4\)
