giải bài toán tìm x biết [55 -x]/1963+[50-x]/1968+[45-x]/1973+[40-x]/1978+4=0 16/10/2021 Bởi Amaya giải bài toán tìm x biết [55 -x]/1963+[50-x]/1968+[45-x]/1973+[40-x]/1978+4=0
Ta có $\dfrac{55-x}{1963} + \dfrac{50-x}{1968} + \dfrac{45-x}{1973}+ \dfrac{40-x}{1978} + 4 = 0$ $<-> \left( \dfrac{55-x}{1963} + 1 \right) + \left( \dfrac{50-x}{1968} + 1 \right) + \left( \dfrac{45-x}{1973} + 1 \right) + \left( \dfrac{40-x}{1978} + 1 \right) = 0$ $<-> \dfrac{2018-x}{1963} + \dfrac{2018-x}{1968} + \dfrac{2018-x}{1973} + \dfrac{2018-x}{1978} = 0$ $<-> (2018-x)\left( \dfrac{1}{1963} + \dfrac{1}{1968} + \dfrac{1}{1973} + \dfrac{1}{1978} \right) = 0$ $<-> 2018-x = 0$ $<-> x = 2018$ Vậy $x = 2018$. Bình luận
Ta có
$\dfrac{55-x}{1963} + \dfrac{50-x}{1968} + \dfrac{45-x}{1973}+ \dfrac{40-x}{1978} + 4 = 0$
$<-> \left( \dfrac{55-x}{1963} + 1 \right) + \left( \dfrac{50-x}{1968} + 1 \right) + \left( \dfrac{45-x}{1973} + 1 \right) + \left( \dfrac{40-x}{1978} + 1 \right) = 0$
$<-> \dfrac{2018-x}{1963} + \dfrac{2018-x}{1968} + \dfrac{2018-x}{1973} + \dfrac{2018-x}{1978} = 0$
$<-> (2018-x)\left( \dfrac{1}{1963} + \dfrac{1}{1968} + \dfrac{1}{1973} + \dfrac{1}{1978} \right) = 0$
$<-> 2018-x = 0$
$<-> x = 2018$
Vậy $x = 2018$.