B=\(\dfrac{1}{6.10}\)+ \(\dfrac{1}{10.14}\)+. . . + \(\dfrac{1}{402.406}\)
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TA
9 tháng 4 2023
Ta có:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{10.11}-\dfrac{1}{11.12}=\dfrac{1}{1.2}-\dfrac{1}{11.12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)Mà \(\dfrac{65}{132}\ne\dfrac{1}{4}\Rightarrow\) Có thể bạn ghi sai đề thì phải !
11 tháng 4 2023
\(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + .....+ \(\dfrac{1}{10.11.12}\)
= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) +....+ \(\dfrac{1}{10.11}\) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{1.2}\) + (- \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\))+.......+ ( \(-\dfrac{1}{10.11}\) + \(\dfrac{1}{10.11}\)) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{11.12}\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{132}\) =\(\dfrac{66}{132}\)-\(\dfrac{1}{132}\) =\(\dfrac{65}{132}\) Vì \(\dfrac{33}{132}\) = \(\dfrac{1}{4}\) nên \(\dfrac{65}{132}\) > \(\dfrac{1}{4}\)
ML
22 tháng 7 2017
\(I=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)
\(\Leftrightarrow I=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\right)\)
\(\Leftrightarrow I=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{406}\right)\)
\(\Leftrightarrow I=\frac{1}{4}\cdot\frac{100}{609}\)
\(\Leftrightarrow I=\frac{25}{609}\)
22 tháng 7 2017
4I = 4/6.10 + 4/10.14 + ......+ 4/402.406 À mình nói thêm tử số sẽ dựa vào khoảng cách giữa 2 mẫu số
= 1/6 -1/10 + 1/10 -1/14+......+1/402 -1/406
= 1/6 - 1/406 = 100/609
I =100/609 : 4 = 25/609
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
DC
4
16 tháng 3 2016
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{402.406}\)
4\(A=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{402}-\frac{1}{406}\)
4\(A=\frac{1}{6}-\frac{1}{406}\)
4\(A=\frac{100}{609}\)
\(\Rightarrow A=\frac{100}{609}:4\)\(=\frac{25}{609}\)
16 tháng 3 2016
=1/6-1/10+1/10-1/14+1/14-1/18+...........+1/402-1/406
=1/6-1/406
17 tháng 5 2022
a: \(P=2x^2+3xy+y^2=\left(2x+y\right)\left(x+y\right)\)
\(=\left(2\cdot\dfrac{-1}{2}+\dfrac{2}{3}\right)\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)\)
\(=\dfrac{-1}{3}\cdot\dfrac{1}{6}=-\dfrac{1}{18}\)
d: \(Q=\dfrac{-1}{3}x^4y^2=\dfrac{-1}{3}\cdot16\cdot\dfrac{1}{16}=-\dfrac{1}{3}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
14 tháng 8 2023
\(=\dfrac{1}{2x1x3x2}+\dfrac{1}{2x2x3x3}+\dfrac{1}{2x3x3x4}+...+\dfrac{1}{2x18x3x19}+\dfrac{1}{2x19x3x20}=\)
\(=\dfrac{1}{2x3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{18x19}+\dfrac{1}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{20-19}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}x\dfrac{19}{20}=\dfrac{19}{120}\)
\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+...+\dfrac{1}{402.406}\\ 4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+...+\dfrac{4}{402.406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\\B=\dfrac{\dfrac{100}{609}}{4}=\dfrac{25}{609} \)
\(B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{ 1}{402}-\dfrac{1}{406}\)
\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}.\)