Toán Tìm x ,biết rằng :1/5.8+1/8.11+…+1/x[x+3]=1/6 05/07/2021 By Arianna Tìm x ,biết rằng :1/5.8+1/8.11+…+1/x[x+3]=1/6
Đặt $A= \frac{1}{5.8}+\frac{1}{8.11}+….+\frac{1}{x.(x+3)}$ ⇔ $3A= \frac{3}{5.8}+\frac{3}{8.11}+….+\frac{3}{x.(x+3)}$ ⇔ $3A= \frac{1}{5}-\frac{1}{x+3}$ ⇔ $A= \frac{x+3-5}{5.(x+3)}:3$ ⇔ $A= \frac{x-2}{15.(x+3)}$ ⇒ $\frac{x-2}{15.(x+3)}= \frac{1}{6}$ ⇔ $6x-12= 15x+45$ ⇔ $9x= -57$ ⇔ $x=\frac{-19}{3}$ ⇒ Không có x thỏa mãn Trả lời
Bạn tham khảo: $\dfrac{1}{5.8} + \dfrac{1}{8.11} + … + \dfrac{1}{x(x+3)} = \dfrac{1}{6}$ $⇔ \dfrac{1}{5} – \dfrac{1}{8} + \dfrac{1}{8} – \dfrac{1}{11} + … + \dfrac{1}{x} – \dfrac{1}{x+3} = \dfrac{1}{6}$ $⇔ \dfrac{1}{5} – \dfrac{1}{x+3} = \dfrac{1}{6}$ $⇔ x = 27$ Trả lời
Đặt $A= \frac{1}{5.8}+\frac{1}{8.11}+….+\frac{1}{x.(x+3)}$
⇔ $3A= \frac{3}{5.8}+\frac{3}{8.11}+….+\frac{3}{x.(x+3)}$
⇔ $3A= \frac{1}{5}-\frac{1}{x+3}$
⇔ $A= \frac{x+3-5}{5.(x+3)}:3$
⇔ $A= \frac{x-2}{15.(x+3)}$
⇒ $\frac{x-2}{15.(x+3)}= \frac{1}{6}$
⇔ $6x-12= 15x+45$
⇔ $9x= -57$
⇔ $x=\frac{-19}{3}$
⇒ Không có x thỏa mãn
Bạn tham khảo:
$\dfrac{1}{5.8} + \dfrac{1}{8.11} + … + \dfrac{1}{x(x+3)} = \dfrac{1}{6}$
$⇔ \dfrac{1}{5} – \dfrac{1}{8} + \dfrac{1}{8} – \dfrac{1}{11} + … + \dfrac{1}{x} – \dfrac{1}{x+3} = \dfrac{1}{6}$
$⇔ \dfrac{1}{5} – \dfrac{1}{x+3} = \dfrac{1}{6}$
$⇔ x = 27$