Số nguyên nhỏ nhất thỏa mãn bất phương trình (2017 - x)/15 + (2018 - x)/16

* Hướng dẫn giải

Đáp án D

Ta có

$$\begin{gathered} \frac{\text{2017}-\text{x}}{\text{15}}+\frac{\text{2018}-\text{x}}{\text{16}}+\frac{\text{17}+\text{x}}{\text{2019}}+\frac{\text{18}+\text{x}}{\text{2020}}\le \text{4} \\ \iff \frac{\text{2017}-\text{x}}{\text{15}}+\frac{\text{2018}-\text{x}}{\text{16}}+\frac{\text{17}+\text{x}}{\text{2019}}+\frac{\text{18}+\text{x}}{\text{2020}}-\text{4}\le \text{0} \\ \iff \frac{\text{2017}-\text{x}}{\text{15}}-1+\frac{\text{2018}-\text{x}}{\text{16}}-1+\frac{\text{17}+\text{x}}{\text{2019}}-1+\frac{\text{18}+\text{x}}{\text{2020}}-1\le \text{0} \\ \iff \frac{\text{2002}-\text{x}}{\text{15}}+\frac{\text{2002}-\text{x}}{\text{16}}+\frac{\text{x-2002}}{\text{2019}}+\frac{\text{x-2002}}{\text{2020}}\le \text{0} \\ \iff -\frac{\text{x-2002}}{\text{15}}-\frac{\text{x-2002}}{\text{16}}+\frac{\text{x-2002}}{\text{2019}}+\frac{\text{x-2002}}{\text{2020}}\le \text{0} \\ \iff (x - 2002)\left(-\frac{\text{1}}{\text{15}}-\frac{\text{1}}{\text{16}}+\frac{\text{1}}{\text{2019}}+\frac{\text{1}}{\text{2020}}\right)\le 0 \end{gathered}$$

Mà $\frac{\text{1}}{\text{2019}}+\frac{\text{1}}{\text{2020}}-\frac{\text{1}}{\text{15}}-\frac{\text{1}}{\text{16}} < 0$  nên

x - 2002 ≥ 0 $\iff$ x ≥ 2002

Vậy giá trị nguyên nhỏ nhất của x là 2002