Tính S = a – b + c – 2d.
Giải thích
\(f'\left( x \right) = \frac{{\left( {3{x^2} + 3} \right)\left( {x - 1} \right) - \left( {{x^3} + 3x + 2} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{2{x^3} - 3{x^2} - 5}}{{{{\left( {x - 1} \right)}^2}}}\);
\(f''\left( x \right) = \frac{{\left( {6{x^2} - 6x} \right){{\left( {x - 1} \right)}^2} - 2\left( {2{x^3} - 3{x^2} - 5} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}}\)\( = \frac{{2{x^3} - 6{x^2} + 6x + 10}}{{{{\left( {x - 1} \right)}^3}}}\).
Suy ra a = 2; b = −6; c = 6; d = 10.
Từ đó a – b + c – 2d = 2 + 6 + 6 – 20 = −6.
Trả lời: −6.