Đáp án:
điều phải chứng minh
Giải thích các bước giải:
\(\begin{array}{l}
y' = \dfrac{{ - 2\left( {x - 3} \right) - \left( {1 - 2x} \right)}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{ - 2x + 6 - 1 + 2x}}{{{{\left( {x - 3} \right)}^2}}} = \dfrac{5}{{{{\left( {x - 3} \right)}^2}}}\\
y'' = \dfrac{{ - 2\left( {x - 3} \right).5}}{{{{\left( {x - 3} \right)}^4}}} = \dfrac{{ - 10\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^4}}}\\
= \dfrac{{ - 10}}{{{{\left( {x - 3} \right)}^3}}}\\
{\left( {y'} \right)^2} = {\left( {\dfrac{5}{{{{\left( {x - 3} \right)}^2}}}} \right)^2} = \dfrac{{25}}{{{{\left( {x - 3} \right)}^4}}}\\
VP = \left( {\dfrac{{1 - 2x}}{{x - 3}} + 2} \right).\dfrac{{ - 10}}{{{{\left( {x - 3} \right)}^3}}}\\
= \dfrac{{1 - 2x + 2x - 6}}{{x - 3}}.\dfrac{{ - 10}}{{{{\left( {x - 3} \right)}^3}}}\\
= \dfrac{{ - 5.\left( { - 10} \right)}}{{{{\left( {x - 3} \right)}^4}}} = \dfrac{{50}}{{{{\left( {x - 3} \right)}^4}}}\\
VT = 2.\dfrac{{25}}{{{{\left( {x - 3} \right)}^4}}} = \dfrac{{50}}{{{{\left( {x - 3} \right)}^4}}}\\
\to VT = VP
\end{array}\)
