\(\frac{x+7}{7x+35}\cdot\frac{x^2-3x-40}{x-8}=(*)\\\\x^2-3x-40=0\\a=1;\ b=-3;\ c=-40\\\Delta=b^2-4ac;\ x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=(-3)^2-4\cdot1\cdot(-40)=9+160=169;\ \sqrt\Delta=\sqrt{169}=13\\\\x_1=\frac{3-13}{2\cdot1}=\frac{-10}{2}=-5;\ x_2=\frac{3+13}{2\cdot1}=\frac{16}{2}=8\\\\x^2-3x-40=(x+5)(x-8)\\\\(*)=\frac{x+7}{7(x+5)}\cdot\frac{(x+5)(x-8)}{x-8}=\frac{x+7}{7}=\frac{x}{7}+\frac{7}{7}=\frac{1}{7}x+1\\\\D:x
eq-5\ \wedge\ x
eq8\)