Nghiệm của bất phương trình dưới đây là: (1987 - x)/15 + (1988 - x)/16 + (27 + x)/1999

* Hướng dẫn giải

Đáp án B

Ta có

$$\begin{gathered} \frac{\text{1987}-\text{x}}{\text{15}}+\frac{\text{1988}-\text{x}}{\text{16}}+\frac{\text{27}+\text{x}}{\text{1999}}+\frac{\text{28}+\text{x}}{\text{2000}}>\text{4} \\ \iff \frac{\text{1987}-\text{x}}{\text{15}}+\frac{\text{1988}-\text{x}}{\text{16}}+\frac{\text{27}+\text{x}}{\text{1999}}+\frac{\text{28}+\text{x}}{\text{2000}}-\text{4}>\text{0} \\ \iff \left(\frac{\text{1987}-\text{x}}{\text{15}}-\text{1}\right)+\left(\frac{\text{1988}-\text{x}}{\text{16}}-\text{1}\right)+\left(\frac{\text{27}+\text{x}}{\text{1999}}-\text{1}\right)+\left(\frac{\text{28}+\text{x}}{\text{2000}}-\text{1}\right)>\text{0} \\ \iff \frac{\text{1972}-\text{x}}{\text{15}}+\frac{\text{1972}-\text{x}}{\text{16}}+\frac{\text{x}-\text{1972}}{\text{1999}}+\frac{\text{x}-\text{1972}}{\text{2000}}>\text{0} \\ \iff (1972 - x)\left(\frac{\text{1}}{\text{15}}+\frac{\text{1}}{\text{16}}-\frac{\text{1}}{\text{1999}}-\frac{\text{1}}{\text{2000}}\right)>0 \end{gathered}$$

Mà $\frac{\text{1}}{\text{15}}+\frac{\text{1}}{\text{16}}-\frac{\text{1}}{\text{1999}}-\frac{\text{1}}{\text{2000}}> 0$ nên 1972 - x > 0

$\iff$ x < 1972

Vậy x < 1972