Toán 1/3+1/6+1/10+….+2/x.(x+1)=2013/2015 find x 08/09/2021 By Adalyn 1/3+1/6+1/10+….+2/x.(x+1)=2013/2015 find x
Đáp án: x = 2014 Giải thích các bước giải: $\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ + … + $\frac{2}{x(x+1)}$ = $\frac{2013}{2015}$ ⇒ $\frac{2}{6}$ + $\frac{2}{12}$ + $\frac{2}{20}$ + … + $\frac{2}{x(x+1)}$ = $\frac{2013}{2015}$ ⇒ $\frac{2}{2.3}$ + $\frac{2}{3.4}$ + $\frac{2}{4.5}$ + … + $\frac{2}{x(x+1)}$ = $\frac{2013}{2015}$ ⇒ 2 . ($\frac{1}{2.3}$ + $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + … + $\frac{1}{x(x+1)}$) = $\frac{2013}{2015}$ ⇒ 2 . ($\frac{1}{2}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – $\frac{1}{4}$ + … + $\frac{1}{x}$ – $\frac{1}{x + 1}$) = $\frac{2013}{2015}$ ⇒ 2 . ($\frac{1}{2}$ – $\frac{1}{x + 1}$) = $\frac{2013}{2015}$ ⇒ $\frac{1}{2}$ – $\frac{1}{x + 1}$ = $\frac{2013}{2015}$ : 2 ⇒ $\frac{1}{2}$ – $\frac{1}{x + 1}$ = $\frac{2013}{2015}$ . $\frac{1}{2}$ ⇒ $\frac{1}{2}$ – $\frac{1}{x + 1}$ = $\frac{2013}{4030}$ ⇒ $\frac{1}{x + 1}$ = $\frac{1}{2}$ – $\frac{2013}{4030}$ ⇒ $\frac{1}{x + 1}$ = $\frac{2015}{4030}$ – $\frac{2013}{4030}$ ⇒ $\frac{1}{x + 1}$ = $\frac{1}{2015}$ ⇒ x + 1 = 2015 ⇒ x = 2015 – 1 ⇒ x = 2014 $Cho mình hay nhất nha$ Trả lời
1/3+1/6+1/10+….+2/x.(x+1)=2013/2015 2/6+2/12+2/20 +….+ 2/x.(x+1) = 2013/2015 2/2×3+2/3×4+2/4×5 +….+ 2/x.(x+1) = 2013/2015 1/2×3+1/2×3+1/3×4+1/3×4+1/4×5+1/4×5+….+ 1/x.(x+1) + 1/x.(x+1) = 2013/2015 1/2 – 1/3 + 1/2 – 1/3 + 1/3 – 1/4 + 1/3 – 1/4 + …. + 1/x – 1/x+1 + 1/x – 1/x+1 = 2013/2015 1/2 – 1/x+1 + 1/2 – 1/x+1 = 2013/2015 1 – 2/x+1 = 2013/2015 2/x+1 = 2/2015 => x= 2014 Trả lời
Đáp án: x = 2014
Giải thích các bước giải:
$\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ + … + $\frac{2}{x(x+1)}$ = $\frac{2013}{2015}$
⇒ $\frac{2}{6}$ + $\frac{2}{12}$ + $\frac{2}{20}$ + … + $\frac{2}{x(x+1)}$ = $\frac{2013}{2015}$
⇒ $\frac{2}{2.3}$ + $\frac{2}{3.4}$ + $\frac{2}{4.5}$ + … + $\frac{2}{x(x+1)}$ = $\frac{2013}{2015}$
⇒ 2 . ($\frac{1}{2.3}$ + $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + … + $\frac{1}{x(x+1)}$) = $\frac{2013}{2015}$
⇒ 2 . ($\frac{1}{2}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – $\frac{1}{4}$ + … + $\frac{1}{x}$ – $\frac{1}{x + 1}$) = $\frac{2013}{2015}$
⇒ 2 . ($\frac{1}{2}$ – $\frac{1}{x + 1}$) = $\frac{2013}{2015}$
⇒ $\frac{1}{2}$ – $\frac{1}{x + 1}$ = $\frac{2013}{2015}$ : 2
⇒ $\frac{1}{2}$ – $\frac{1}{x + 1}$ = $\frac{2013}{2015}$ . $\frac{1}{2}$
⇒ $\frac{1}{2}$ – $\frac{1}{x + 1}$ = $\frac{2013}{4030}$
⇒ $\frac{1}{x + 1}$ = $\frac{1}{2}$ – $\frac{2013}{4030}$
⇒ $\frac{1}{x + 1}$ = $\frac{2015}{4030}$ – $\frac{2013}{4030}$
⇒ $\frac{1}{x + 1}$ = $\frac{1}{2015}$
⇒ x + 1 = 2015
⇒ x = 2015 – 1
⇒ x = 2014
$Cho mình hay nhất nha$
1/3+1/6+1/10+….+2/x.(x+1)=2013/2015
2/6+2/12+2/20 +….+ 2/x.(x+1) = 2013/2015
2/2×3+2/3×4+2/4×5 +….+ 2/x.(x+1) = 2013/2015
1/2×3+1/2×3+1/3×4+1/3×4+1/4×5+1/4×5+….+ 1/x.(x+1) + 1/x.(x+1) = 2013/2015
1/2 – 1/3 + 1/2 – 1/3 + 1/3 – 1/4 + 1/3 – 1/4 + …. + 1/x – 1/x+1 + 1/x – 1/x+1 = 2013/2015
1/2 – 1/x+1 + 1/2 – 1/x+1 = 2013/2015
1 – 2/x+1 = 2013/2015
2/x+1 = 2/2015
=> x= 2014