A=5/4x7 + 5/7x10 + ........ + 5/25x28 + 5/28x31
nhanh hộ e nhé mn
bài tính nhanh
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HH
2
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
\(\dfrac{5}{4\times7}+\dfrac{5}{7\times10}+....+\dfrac{5}{25\times28}+\dfrac{5}{28\times31}\)
\(=5\times\left(\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{28\times31}\right)\)
\(=\dfrac{5}{3}\times\left(\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+....+\dfrac{3}{28\times31}\right)\)
\(=\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{28}-\dfrac{1}{31}\right)\)
\(=\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{31}\right)\)
\(=\dfrac{5}{3}\times\dfrac{27}{124}\)
\(=\dfrac{45}{124}\)
24 tháng 9 2023
`#3107`
\(\dfrac{5}{4\times7}+\dfrac{5}{7\times10}+...+\dfrac{5}{25\times28}+\dfrac{5}{28\times31}\)
\(=\dfrac{5}{3}\times\left(\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{25\times28}+\dfrac{3}{28\times31}\right)\)
\(=\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{28}-\dfrac{1}{31}\right)\\ =\dfrac{5}{3}\times\left(\dfrac{1}{4}-\dfrac{1}{31}\right)\\ =\dfrac{5}{3}\times\dfrac{27}{124}\\ =\dfrac{45}{124}\)
LM
Tính nhanh
a) 5/1x3 + 5/3x5 + 5/5x7 + ........ + 5/43x45
b) 6/1x4 + 6/4x7 + 6/7x10 + ...... + 6/97x100
1
DN
21 tháng 8 2015
\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)
AH
Akai Haruma
Giáo viên
6 tháng 12 2023
Bài 1:
$M=3.4.5+4.5.6+...+13.14.15$
$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$
$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$
$=-2.3.4.5+13.14.15.16=43560$
$M=43560:4=10890$
AH
Akai Haruma
Giáo viên
6 tháng 12 2023
Bài 2:
a.
$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$
$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$
$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
$M=\frac{99}{100}:3=\frac{33}{100}$
MA
1
22 tháng 5 2021
Bài 1:
b) \(B=A.\dfrac{-10}{x-4}=\dfrac{x-4}{x+5}.\dfrac{-10}{x-4}=\dfrac{-10}{x+5}\)
Để B nguyên <=> x+5 nguyên mà \(x\in Z\Rightarrow x+5\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
\(\Leftrightarrow x\in\left\{-6;-4;-3;-7;0;-10;-15;5\right\}\) kết hợp với điều kiện của x
\(\Rightarrow x\in\left\{-15;-10;-6;-7;-3;0;5\right\}\)
Bài 5:
Có \(\left|x-2018\right|+\left|2x-2019\right|+\left|3x-2020\right|\ge0\) \(\forall\)x
\(\Rightarrow x-2021\ge0\) \(\Leftrightarrow x\ge2021\)
\(\Rightarrow x-2018>0,2x-2019>0,3x-2020>0\)
PT \(\Leftrightarrow x-2018+2x-2019+3x-2020=x-2021\)
\(\Leftrightarrow5x=4036\) \(\Leftrightarrow x=\dfrac{4036}{5}< 2021\) (L)
Vậy pt vô nghiệm
NT
5
Z
18 tháng 7 2019
Ta có:
A = \(\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{301.304}\)
A = 5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))
3A = 3.5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))
3A = 5. (\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{304}\))
3A = \(\frac{5.75}{304}\)
3A = \(\frac{375}{304}\)
A= \(\frac{125}{304}\) . Vậy: A = \(\frac{125}{304}\)
Chúc bạn học tốt!
Tick cho mình nhé!
DG
1
S
23 tháng 9 2023
\(E=\dfrac{1}{1\times2}+\dfrac{2}{2\times4}+\dfrac{3}{4\times7}+\dfrac{4}{7\times11}+\dfrac{5}{11\times16}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)
\(=1-\dfrac{1}{16}=\dfrac{15}{16}\)
#kễnh
\(A=\dfrac{5}{4x7}+\dfrac{5}{7x10}+...+\dfrac{5}{25x28}+\dfrac{5}{28x31}\)
\(\dfrac{3}{5}A=\dfrac{7-4}{4x7}+\dfrac{10-7}{7x10}+...+\dfrac{28-25}{25x28}+\dfrac{31-28}{28x31}\)
\(\dfrac{3}{5}A=\dfrac{7}{4x7}-\dfrac{4}{4x7}+\dfrac{10}{7x10}-\dfrac{7}{7x10}+...+\dfrac{28}{25x28}-\dfrac{25}{25x28}+\dfrac{31}{28x31}-\dfrac{28}{28x31}\)
\(\dfrac{3}{5}A=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}+\dfrac{1}{28}-\dfrac{1}{31}\)
\(\dfrac{3}{5}A=\dfrac{1}{4}-\dfrac{1}{31}=\dfrac{27}{124}\)
\(A=\dfrac{27}{124}:\dfrac{3}{5}=\dfrac{27}{124}x\dfrac{5}{3}=\dfrac{45}{124}\)