Tìm M = \(\dfrac{20}{112}+\dfrac{20}{180}+\dfrac{20}{520}+\dfrac{20}{832}\)
Thực hiện phép tính sau:
M=\(\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\\ =\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\\ =\dfrac{20}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\\ =\dfrac{20}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\\ =\dfrac{20}{6}.\left(\dfrac{1}{8}-\dfrac{1}{32}\right)\\ =\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
Thực hiện phép tính sau:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
Anh giải cách đây 3 ngày rồi!
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\\ M=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\\ M=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\\M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\\ M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(\Rightarrow M=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(\Rightarrow M=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{8}{14.20}+\dfrac{8}{20.26}+\dfrac{8}{26.32}\right)\)
\(\Rightarrow M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(\Rightarrow M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)\)
\(\Rightarrow M=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(M=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(M=\dfrac{20}{6}.\left(\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\right)\)
\(M=\dfrac{20}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(M=\dfrac{20}{6}.\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{6}{20}.\dfrac{3}{32}=\dfrac{5}{16}\)
Chúc bạn học tốt!!!
4. tính
M= \(\dfrac{20}{112}\)+\(\dfrac{20}{280}\)+\(\dfrac{20}{520}\)+\(\dfrac{20}{832}\)
Ta có :
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(M=\dfrac{20}{8\times14}+\dfrac{20}{14\times20}+\dfrac{20}{20\times26}+\dfrac{20}{26\times32}\)
\(\Rightarrow\dfrac{3}{10}M=\dfrac{6}{8\times14}+\dfrac{6}{14\times20}+\dfrac{6}{20\times26}+\dfrac{6}{26\times32}\)
\(\dfrac{3}{10}M=\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\)
\(\dfrac{3}{10}M=\dfrac{1}{8}-\dfrac{1}{32}=\dfrac{3}{32}\)
\(\Rightarrow M=\dfrac{3}{32}\div\dfrac{3}{10}=\dfrac{5}{16}\)
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(M=20.\left(\dfrac{1}{112}+\dfrac{1}{280}+\dfrac{1}{520}+\dfrac{1}{832}\right)\)
\(M=20.\left(\dfrac{1}{8.14} +\dfrac{1}{14.20}+\dfrac{1}{20.26}+\dfrac{1}{26.32}\right)\)
\(\Rightarrow6M=20.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)
\(\Rightarrow6M=20.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(\Rightarrow6M=20.\left(\dfrac{1}{8}-\dfrac{1}{32}\right)\)
\(\Rightarrow6M=20.\dfrac{3}{32}\)
\(\Rightarrow6M=\dfrac{15}{8}\)
\(\Rightarrow M=\dfrac{15}{8}:6\)
\(\Rightarrow M=\dfrac{5}{16}\)
Vậy \(M=\dfrac{5}{16}\)
Thực hiện phép tính sau:
M=\(\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
Tính tổng
A=\(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}+\dfrac{1}{182}+\dfrac{1}{210}\)
Bài 1: Ta có:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
Vậy \(M=\dfrac{5}{16}\)
Bài 2: Ta có:
\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)
\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)
\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)
Vậy \(A=\dfrac{1}{10}\)
Giải:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)
\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)
\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)
\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)
\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)
M=\(\dfrac{5}{28}+\dfrac{1}{14}+\dfrac{1}{26}+\dfrac{5}{208}\)
M=\((\dfrac{5}{28}+\dfrac{1}{14})+\left(\dfrac{1}{26}+\dfrac{5}{208}\right)\)
M=\(\dfrac{1}{4}+\dfrac{1}{16}\)
M=\(\dfrac{5}{16}\)
A=\(\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+...+\dfrac{1}{13\times14}+\dfrac{1}{14\times15}\)
A=\(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)
Sau khi giản ước các phân số cho nhau Ta có:
A=\(\dfrac{1}{6}-\dfrac{1}{15}\)
A=\(\dfrac{1}{10}\)
M=20/112+20/180+20/520+20/832 = ?
M=20/112+20/280+20/520+20/832
M = 5/16 = 0,3125
Tích cho tớ để đạt 250 điểm nha xin đó
Tính M=20/112+20/280+20/520+20/832
M=20/8.14+20/14.20+20/20.26+20/26.32
=>M=20/6(6/8.14+6/14.20+6/20.26+6/26.32)
=>M=20/6(1/8-1/14+1/14-1/20+1/20-1/26+1/26-1/32)
=>M=20/6(1/8-1/32)
= 20/6.3/32
=5/16
Tính M,biết: M=20/112+20/280+20/520+20/832
\(M=\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}\)
\(M=\frac{20}{8.14}+\frac{20}{14.20}+\frac{20}{20.26}+\frac{20}{26.32}\)
\(M=\frac{20}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+\frac{6}{26.32}\right)\)
\(M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+\frac{1}{26}-\frac{1}{32}\right)\)
\(M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{32}\right)\)
\(M=\frac{20}{6}\cdot\frac{3}{32}\)
\(M=\frac{5}{16}\)
$M=\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}$
$M=\frac{20}{8.14}+\frac{20}{14.20}+\frac{20}{20.26}+\frac{20}{26.32}$
$M=\frac{20}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+\frac{6}{26.32}\right)$
$M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+\frac{1}{26}-\frac{1}{32}\right)$
$M=\frac{20}{6}\left(\frac{1}{8}-\frac{1}{32}\right)$
$M=\frac{20}{6}\cdot \frac{3}{32}$
$M=\frac{5}{16}$
tính nhanh:
M=20/112+20/280+20/520+20/832