Bài tập 5.52 trang 208 SBT Toán 11

Ta có:

\(\begin{array}{l}
y' = {\left( {x - 2} \right)^2}{\left( {x - 3} \right)^3} + \left( {x - 1} \right)\left[ {2\left( {x - 2} \right){{\left( {x - 3} \right)}^3} + {{\left( {x - 2} \right)}^2}.3{{\left( {x - 3} \right)}^2}} \right]\\
 = {\left( {x - 2} \right)^2}{\left( {x - 3} \right)^3} + \left( {x - 1} \right)\left( {x - 2} \right){\left( {x - 3} \right)^2}\left[ {2\left( {x - 3} \right) + 3\left( {x - 2} \right)} \right]\\
 = {\left( {x - 2} \right)^2}{\left( {x - 3} \right)^2} + \left( {x - 1} \right)\left( {x - 2} \right){\left( {x - 3} \right)^2}\left( {5x - 12} \right)\\
 = (x - 2){(x - 3)^2}({x^2} - 5x + 6 + 5{x^2} - 17x + 12)\\
 = (x - 2){(x - 3)^2}(6{x^2} - 22x + 18)
\end{array}\)

Suy ra: 

\(f'(1) = (1 - 2){(1 - 3)^2}(6 - 22 + 18) = 8\)

\(\begin{array}{l}
f'(2) = (2 - 2){(2 - 3)^2}(6.4 - 22.2 + 18) = 0\\
f'(3) = (3 - 2){(3 - 3)^2}(6.9 - 22.3 + 18) = 0
\end{array}\)

-- Mod Toán 11 HỌC247