Toán Xét chiều biến thiên của hàm số: y=x^2-3x+1 trên (2;+ vô cùng) 18/09/2021 By Eden Xét chiều biến thiên của hàm số: y=x^2-3x+1 trên (2;+ vô cùng)
$\begin{array}{l} y = {x^2} – 3x + 1\\ Voi\,{x_1},{x_2} \in \left( {2; + \infty } \right)\,ta\,co:\\ \frac{{f\left( {{x_1}} \right) – f\left( {{x_2}} \right)}}{{{x_1} – {x_2}}} = \frac{{x_1^2 – 3{x_1} + 1 – x_2^2 – 3{x_2} + 1}}{{{x_1} – {x_2}}}\\ = \frac{{\left( {x_1^2 – x_2^2} \right) – 3\left( {{x_1} – {x_2}} \right)}}{{{x_1} – {x_2}}} = \frac{{\left( {{x_1} + {x_2}} \right)\left( {{x_1} – {x_2}} \right) – 3\left( {{x_1} – {x_2}} \right)}}{{{x_1} – {x_2}}}\\ = \frac{{\left( {{x_1} – {x_2}} \right)\left( {{x_1} + {x_2} – 3} \right)}}{{{x_1} – {x_2}}} = {x_1} + {x_2} – 3\\ Vi\,{x_1},{x_2} > 2 \Rightarrow {x_1} + {x_2} – 3 > 0 \Rightarrow hs\,dong\,bien\,tren\,\left( {2; + \infty } \right) \end{array}$ Trả lời
$\begin{array}{l}
y = {x^2} – 3x + 1\\
Voi\,{x_1},{x_2} \in \left( {2; + \infty } \right)\,ta\,co:\\
\frac{{f\left( {{x_1}} \right) – f\left( {{x_2}} \right)}}{{{x_1} – {x_2}}} = \frac{{x_1^2 – 3{x_1} + 1 – x_2^2 – 3{x_2} + 1}}{{{x_1} – {x_2}}}\\
= \frac{{\left( {x_1^2 – x_2^2} \right) – 3\left( {{x_1} – {x_2}} \right)}}{{{x_1} – {x_2}}} = \frac{{\left( {{x_1} + {x_2}} \right)\left( {{x_1} – {x_2}} \right) – 3\left( {{x_1} – {x_2}} \right)}}{{{x_1} – {x_2}}}\\
= \frac{{\left( {{x_1} – {x_2}} \right)\left( {{x_1} + {x_2} – 3} \right)}}{{{x_1} – {x_2}}} = {x_1} + {x_2} – 3\\
Vi\,{x_1},{x_2} > 2 \Rightarrow {x_1} + {x_2} – 3 > 0 \Rightarrow hs\,dong\,bien\,tren\,\left( {2; + \infty } \right)
\end{array}$