回復 16# mathca 的帖子
第5題，從你的 \( b+c=4 \) 硬做就好了，移項平方兩次
\( \sqrt{(x-1)^{2}+\frac{27}{25}}+\sqrt{(x+1)^{2}+\frac{27}{25}}=4 \)
\(\Rightarrow\left[\sqrt{(x-1)^{2}+\frac{27}{25}}\right]^{2}=\left[4-\sqrt{(x+1)^{2}+\frac{27}{25}}\right]^{2} \)
\( \Rightarrow(x-1)^{2}+\frac{27}{25}=16+(x+1)^{2}+\frac{27}{25}-8\sqrt{(x+1)^{2}+\frac{27}{25}} \)
\( \Rightarrow-4x-16=-8\sqrt{(x+1)^{2}+\frac{27}{25}} \)
\( \Rightarrow(x+4)^{2}=4\left[(x+1)^{2}+\frac{27}{25}\right] \)
\( \Rightarrow3x^{2}-\frac{192}{25}=0
\Rightarrow x=\pm\frac{8}{5} \)
又 \( b \geq c \) 故 \( x = \frac{8}{5} \)
