# Bài 2 trang 156 SGK Đại số và Giải tích 11

##### Hướng dẫn giải

Tính $$\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)$$, từ đó tính tỉ số $$\frac{{\Delta y}}{{\Delta x}}$$.

Lời giải chi tiết

$$\begin{array}{l} a)\,\,\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\\ \Rightarrow \Delta y = 2\left( {x + \Delta x} \right) - 5 - \left( {2x - 5} \right)\\ \Leftrightarrow \Delta y = 2x + 2\Delta x - 5 - 2x + 5\\ \Leftrightarrow \Delta y = 2\Delta x\\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = 2\\ b)\,\,\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\\ \Rightarrow \Delta y = {\left( {x + \Delta x} \right)^2} - 1 - \left( {{x^2} - 1} \right)\\ \Leftrightarrow \Delta y = {x^2} + 2x.\Delta x + {\left( {\Delta x} \right)^2} - 1 - {x^2} + 1\\ \Leftrightarrow \Delta y = \Delta x\left( {2x + \Delta x} \right)\\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = 2x + \Delta x\\ c)\,\,\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\\ \Rightarrow \Delta y = 2{\left( {x + \Delta x} \right)^3} - 2{x^3}\\ \Leftrightarrow \Delta y = 2{x^3} + 6{x^2}\Delta x + 6x{\left( {\Delta x} \right)^2} + 2{\left( {\Delta x} \right)^3} - 2{x^3}\\ \Leftrightarrow \Delta y = 2\Delta x\left( {3{x^2} + 3x.\Delta x + {{\left( {\Delta x} \right)}^2}} \right)\\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = 2\left( {3{x^2} + 3x.\Delta x + {{\left( {\Delta x} \right)}^2}} \right)\\ d)\,\,\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\\ \Rightarrow \Delta y = \frac{1}{{x + \Delta x}} - \frac{1}{x}\\ \Leftrightarrow \Delta y = \frac{{x - x - \Delta x}}{{x\left( {x + \Delta x} \right)}}\\ \Leftrightarrow \Delta y = \frac{{ - \Delta x}}{{x\left( {x + \Delta x} \right)}}\\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{{ - 1}}{{x\left( {x + \Delta x} \right)}} \end{array}$$