\(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)
<=> \(\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)
<=> \(\left|x-4\right|+\left|x+2\right|=0\)
<=> \(\left|4-x\right|+\left|x+2\right|=0\)
Ta thấy: \(\left|4-x\right|+\left|x+2\right|\ge\left|4-x+x+2\right|=\left|6\right|=6\)
mà \(\left|4-x\right|+\left|x+2\right|=0\)
=> pt vô nghiệm