Bài 1:Tìm x: a,1/5.8+1/8.11+1/11.14+…1/x.(x+3)=101/1540 14/10/2021 Bởi Ayla Bài 1:Tìm x: a,1/5.8+1/8.11+1/11.14+…1/x.(x+3)=101/1540
Đáp án: Giải thích các bước giải: $\dfrac{1}{5.8}+\dfrac{1}{8.11}+…+\dfrac{1}{x.(x+3)}=\dfrac{101}{1540}$ $⇔\dfrac{1}{3}.(\dfrac{3}{5.8}+\dfrac{3}{8.11}+…+\dfrac{3}{x.(x+3)}=\dfrac{101}{1540}$ $ $ $⇔\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+…+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}$ $ $ $⇔\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}$ $ $ $⇔\dfrac{1}{x+3}=\dfrac{1}{308}$ $ $ $⇒x+3=308$ $⇒x=305$ Bình luận
$\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+…+\frac{1}{x(x+3)}=\frac{101}{1540}$ $⇔\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+…+\frac{3}{x(x+3)}=\frac{303}{1540}$ $⇔\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+…+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}$ $⇔\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}$ $⇔\frac{1}{x+3}=\frac{1}{308}$ $⇔x+3=308$ $⇔x=305$ Vậy $x=305$. Bình luận
Đáp án:
Giải thích các bước giải:
$\dfrac{1}{5.8}+\dfrac{1}{8.11}+…+\dfrac{1}{x.(x+3)}=\dfrac{101}{1540}$
$⇔\dfrac{1}{3}.(\dfrac{3}{5.8}+\dfrac{3}{8.11}+…+\dfrac{3}{x.(x+3)}=\dfrac{101}{1540}$
$ $
$⇔\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+…+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}$
$ $
$⇔\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}$
$ $
$⇔\dfrac{1}{x+3}=\dfrac{1}{308}$
$ $
$⇒x+3=308$
$⇒x=305$
$\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+…+\frac{1}{x(x+3)}=\frac{101}{1540}$
$⇔\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+…+\frac{3}{x(x+3)}=\frac{303}{1540}$
$⇔\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+…+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}$
$⇔\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}$
$⇔\frac{1}{x+3}=\frac{1}{308}$
$⇔x+3=308$
$⇔x=305$
Vậy $x=305$.