Question

\(x-4+\sqrt{x^2}+8x+16\) với x←4

Answer 1

\(x-4+\sqrt{x^2}+8x+16\left(1\right)\)

TH1: \(0\le x< 4\)

\(\left(1\right)=x-4+x+8x+16=2x+8x+12\)

TH2: \(x< 0\)

\(\left(1\right)=x-4-x+8x+16=8x+12\)