Tính các biểu thức sau
a,2/1x3x5+2/3x5x7+2/5x7x9+2/7x9x11
b,1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6
K
Khách
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Những câu hỏi liên quan
TH
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
A = \(\dfrac{2}{1\times3\times5}\) + \(\dfrac{2}{3\times5\times7}\) + \(\dfrac{2}{5\times7\times9}\)+\(\dfrac{2}{7\times9\times11}\)
A = \(\dfrac{1}{2}\) x (\(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) + \(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+\(\dfrac{1}{5\times7}\)-\(\dfrac{1}{7\times9}\)+\(\dfrac{1}{7\times9}\)-\(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\) - \(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{99}\))
A = \(\dfrac{1}{2}\times\) \(\dfrac{32}{99}\)
A = \(\dfrac{16}{99}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
B = \(\dfrac{1}{1\times2\times3}\) + \(\dfrac{1}{2\times3\times4}\) + \(\dfrac{1}{3\times4\times5}\) + \(\dfrac{1}{4\times5\times6}\)
B = \(\dfrac{1}{2}\) x (\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}\))
B = \(\dfrac{1}{2}\) x (\(\dfrac{1}{1\times2}\)-\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times3}\)-\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{3\times4}\)-\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{4\times5}\)-\(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x(\(\dfrac{1}{1\times2}\) - \(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}-\dfrac{1}{30}\))
B = \(\dfrac{1}{2}\)x \(\dfrac{7}{15}\)
B = \(\dfrac{7}{30}\)
1 tháng 3 2018
2A=\(\frac{2}{1.2.3}\)+\(\frac{2}{2.3.4}\)+...+\(\frac{2}{18.19.20}\)
=1/1.2-1/2.3+1/2.3-1/3.4+...+1/18.19-1/19.20
=1/2-1/19.20
A=1/4-1/19.20.2
vậy A<1/4
VK
6
5 tháng 8 2016
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{1.2}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{56}\)
\(=\frac{28}{56}-\frac{1}{56}=\frac{27}{56}\)
Dấu . là nhân nha
5 tháng 8 2016
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)
\(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)
.......................................
\(\frac{2}{6.7.8}=\frac{1}{6.7}-\frac{1}{7.8}\)
S= \(\frac{1}{1.2}-\frac{1}{7.8}=\frac{27}{56}\)
KA
0
MT
1
31 tháng 1 2021
Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)
\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)
hay \(A=\dfrac{-4949}{19800}\)
PM
1
PN
24 tháng 2 2016
\(\text{Charlotte :'(}\)
Giải phương trình.
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{637}{2550}\) \(\left(\text{*}\right)\)
\(ĐKXĐ:\) \(x\ne0;\) \(x\ne-1;\) và \(x\ne-2\)
Ta có:
\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)
\(.....................\)
\(\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{1}{2}\left(\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\)
Khi đó, phương trình \(\left(\text{*}\right)\) tương đương với
\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{637}{2550}\)
\(\Leftrightarrow\) \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{637}{2550}\)
\(\Leftrightarrow\) \(\frac{1}{4}-\frac{1}{2\left(x+1\right)\left(x+2\right)}=\frac{637}{2550}\)
\(\Leftrightarrow\) \(\frac{1}{2\left(x+1\right)\left(x+2\right)}=\frac{1}{5100}\)
\(\Rightarrow\) \(2\left(x+1\right)\left(x+2\right)=5100\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(x+2\right)=2550\)
\(\Leftrightarrow\) \(^{x_1=-52}_{x_2=49}\) (t/m điều kiện xác định)
Vậy, tập nghiệm của pt \(\left(\text{*}\right)\) là \(S=\left\{-52;49\right\}\)
NM
1
AH
Akai Haruma
Giáo viên
28 tháng 7
Lời giải:
$4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+...+k(k+1)(k+2)[(k+3)-(k-1)]$
$=[1.2.3.4+2.3.4.5+3.4.5.6+...k(k+1)(k+2)(k+3)]-[0.1.2.3+1.2.3.4+2.3.4.5+....+(k-1)k(k+1)(k+2)]$
$=k(k+1)(k+2)(k+3)$
$\Rightarrow 4S+1=k(k+1)(k+2)(k+3)+1=[k(k+3)][(k+1)(k+2)]+1$
$=(k^2+3k)(k^2+3k+2)+1=(k^2+3k)^2+2(k^2+3k)+1=(k^2+3k+1)^2$
$\Rightarrow 4S+1$ là số chính phương.