Ta có:
\(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(\Rightarrow\left(3x-9\right).\left(x+7\right)+x^2-4.2.x+4^2+48\)
\(\Rightarrow3x\left(x+7\right)-9\left(x+7\right)+x^2-8x+16+48\)
\(\Rightarrow9x^2+21x-9x+63+x^2-8x+64\)
\(\Rightarrow\left(9x^2+x^2\right)+\left(21x-9x-8x\right)+63+64=10x^2+4x+127\)
Tại x = 0,5
\(\Rightarrow10.0,5^2+4.0,5+127=131,5\)