tìm x biết 1/5.8+1/8.11+1/11.14+…+1/x+(x+3)=134/2015 tìm x, y thuộc z biết 2y/3 – 5/x+1=1/3 18/11/2021 Bởi Rose tìm x biết 1/5.8+1/8.11+1/11.14+…+1/x+(x+3)=134/2015 tìm x, y thuộc z biết 2y/3 – 5/x+1=1/3
a) $\dfrac{1}{5.8}+\dfrac{1}{8.11}+….+\dfrac{1}{x.(x+3)}=\dfrac{134}{2015}$ $\to \dfrac{1}{3}.\bigg(\dfrac{3}{5.8}+\dfrac{3}{8.11}+….+\dfrac{1}{x.(x+3)} \bigg) = \dfrac{134}{2015}$ $\to \dfrac{1}{5} – \dfrac{1}{8}+\dfrac{1}{8} – …..+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{402}{2015}$ $\to \dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{402}{2015}$ $\to \dfrac{x-2}{x+3} = \dfrac{402}{403}$ $\to 403x-806=402x+1206$ $\to x = 2012$ Vậy $x=2012$ b)$ \dfrac{2y}{3}-\dfrac{5}{x+1}=\dfrac{1}{3}$ $\to \dfrac{5}{x+1}=\dfrac{2y-1}{3}$ $\to (x+1).(2y-1)=15$ Đến đây em lập bảng ra xét số nguyên $x,y$ nhé ! Bình luận
a) $\dfrac{1}{5.8}+\dfrac{1}{8.11}+….+\dfrac{1}{x.(x+3)}=\dfrac{134}{2015}$
$\to \dfrac{1}{3}.\bigg(\dfrac{3}{5.8}+\dfrac{3}{8.11}+….+\dfrac{1}{x.(x+3)} \bigg) = \dfrac{134}{2015}$
$\to \dfrac{1}{5} – \dfrac{1}{8}+\dfrac{1}{8} – …..+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{402}{2015}$
$\to \dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{402}{2015}$
$\to \dfrac{x-2}{x+3} = \dfrac{402}{403}$
$\to 403x-806=402x+1206$
$\to x = 2012$
Vậy $x=2012$
b)$ \dfrac{2y}{3}-\dfrac{5}{x+1}=\dfrac{1}{3}$
$\to \dfrac{5}{x+1}=\dfrac{2y-1}{3}$
$\to (x+1).(2y-1)=15$
Đến đây em lập bảng ra xét số nguyên $x,y$ nhé !