\(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2}{8}-\frac{16\left(2x+1\right)}{8}=\frac{200}{8}+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\left(x+2\right)^2-32x-16=200+\left(x-2\right)^2\)
\(\Leftrightarrow x^2+4x+4-32x-16-200=x^2-4x+4\)
\(\Leftrightarrow x^2-28x-212-x^2+4x-4=0\)
\(\Leftrightarrow-24x=216\)
\(\Leftrightarrow x=-9\)
TL:
a)
\(\frac{\left(x+2\right)^2}{8}-\frac{16\left(2x+1\right)}{8}=\frac{200+\left(x-2\right)^2}{8}\)
\(\frac{x^2+4x+4-32x-16}{8}=\frac{200+x^2-4x+4}{8}\)
\(x^2-28x-12-200-x^2+4x-4=0\)
\(-24x-216=0\)
\(-24x=216\)
\(x=-9\)
Vậy x=-9
\(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
\(\Leftrightarrow\frac{3\left(4x^2-9\right)}{24}=\frac{4\left(x-4\right)^2}{24}+\frac{8\left(x-2\right)^2}{24}\)
\(\Leftrightarrow12x^2-27=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\)
\(\Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\)
\(\Leftrightarrow12x^2-27=12x^2-64x+96\)
\(\Leftrightarrow64x=123\)
\(\Leftrightarrow x=\frac{123}{64}\)