1/1×2 + 1/2×3 + 1/3×4 +......+ 1/99×100
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20 tháng 12 2022
\(a,\left|x+2\right|=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
\(b,\left|x-5\right|=\left|-7\right|\)
\(\Leftrightarrow\left|x-5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
\(c,\left(7-x\right)-\left(25+7\right)=-25\)
\(\Leftrightarrow7-x-32=-25\)
\(\Leftrightarrow x=0\)
\(d,\left|x-3\right|=\left|5\right|+\left|-7\right|\)
\(\Leftrightarrow\left|x-3\right|=12\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
PK
1
H
1 tháng 9 2021
\(B=1-5+5^{^2}-5^{^3}+...-5^{^{99}}+5^{^{100}}\)
\(5B=5-5^{^2}+5^{^3}-5^{^4}+...-5^{^{100}}+5^{^{101}}\)
\(5B+B=\left(5-5^{^2}+5^{^3}-5^{^4}+...-5^{^{100}}+5^{^{101}}\right)+\left(1-5+5^{^2}-5^{^3}+...-5^{^{99}}+5^{^{100}}\right)\)
\(6B=5^{^{101}}+1\)
\(B=\dfrac{5^{^{101}}+1}{6}\)
TA
5 tháng 1 2016
= 1
tick đi mink giải thích cho . hihihihihihihihiihihiiiiiiiiiiiiiiii
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}\)+ \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)
A = 1
CG
2
20 tháng 12 2015
\(a_{n-1}=\frac{1}{1+2+..+n}=\frac{2}{n\left(n+1\right)}=\frac{2}{n}-\frac{2}{n+1}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{99}-\frac{2}{100}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
D
17 tháng 6 2018
\(1\times\left(1+1\right)+2\times\left(2+1\right)+3\times\left(3+1\right)\)
\(=1\times2+2\times3+3\times4\)
\(=2+6+12\)
\(=20\)
\(a=215\times62+42-52\times215\)
\(a=215\times\left(62-52\right)+42\)
\(a=215\times10+42\)
\(a=2150+42\)
\(a=2192\)
\(b=14\times29+14\times71+\left(1+2+3+...+99\right)\times\left(199199\times198-198198\times199\right)\)
\(b=14\times\left(29+71\right)+\left(1+2+3+...+99\right)\times\left(199\times1001\times198-198\times1001\times199\right)\)
\(b=14\times100+0\)
\(b=1400\)
17 tháng 6 2018
1: Quá dễ
1 . (1 + 1) + 2 . (2 + 1) + 3 . (3 + 1)
= 1 . 2 + 2 . 3 + 3 . 4
= 2 + 6 + 12
= 20
2:
a = 215 . 62 + 42 - 52 . 215
= 215 . (62 - 52) + 42
= 215 . 10 + 42
= 2150 + 42
= 2192
b = 14 . 29 + 14 . 71 + (1 + 2 + 3 + ... + 99) . (199199 . 198 - 198198 . 199)
= 14 . (29 + 71) + (1 + 2 + 3 + ... + 99) . (199 . 1001 . 198 - 198 . 1001 . 199)
= 14 . 100 + (1 + 2 + 3 + ... + 99) . 0
= 1400 + 0 = 1400
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=1.\frac{99}{100}\)
\(=\frac{99}{100}\)
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