Tính
\(3\frac{1}{1117}.\frac{4}{1119}-1\frac{1116}{1117}.5\frac{1118}{1119}-\frac{5}{1119}\)
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TN
1
24 tháng 7 2017
Đặt A=\(\frac{9}{10}+\frac{39}{40}+...+\frac{1119}{1120}\)
=>A=\(\frac{10-1}{10}+\frac{40-1}{40}+...+\frac{1120-1}{1120}\)
=>A=\(1-\frac{1}{10}+1-\frac{1}{40}+...+1-\frac{1}{1120}\)
=>A=\(11-\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\right)\)
Đặt B=\(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\)
=>3B=\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{1120}\)
=>3B=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)
=>3B=\(\frac{33}{70}\)
=>B=\(\frac{11}{70}\)
=>A=11-\(\frac{11}{70}\)
=>A=\(\frac{759}{70}\)
BM
0
M
9 tháng 4 2018
1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)
=4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)
= 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)
=4-\(\left(1-\frac{1}{9}\right)\)
= 4-\(\frac{8}{9}\)
= \(\frac{7}{9}\)
14 tháng 4 2018
Ta có :
\(\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+...+\frac{1119}{1120}\)
\(=\)\(\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{40}\right)+\left(1-\frac{1}{88}\right)+...+\left(1-\frac{1}{1120}\right)\)
\(=\)\(\left(1+1+1+...+1\right)-\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{1120}\right)\)
\(=\)\(\left(1+1+1+...+1\right)-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{32.35}\right)\)
\(=\)\(11-\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{32}-\frac{1}{35}\right)\)
\(=\)\(11-\left(\frac{1}{2}-\frac{1}{35}\right)\)
\(=\)\(11-\frac{33}{70}\)
\(=\)\(\frac{737}{70}\)
Chúc bạn học tốt ~
NN
0
NN
1
AH
Akai Haruma
Giáo viên
2 tháng 12 2023
Lời giải:
$D=79(1119+59)-1119.29-179.59$
$=79.1119+79.59-1119.29-179.59$
$=(79.1119-1119.29)-(179.59-79.59)$
$=1119(79-29)-59(179-79)$
$=1119.50-59.100=55950-5900=50050$
\(3\frac{1}{1117}.\frac{4}{1119}-1\frac{1116}{1117}.5\frac{1118}{1119}-\frac{5}{1119}\)
\(=\frac{3352}{1117}.\frac{4}{1119}-\frac{2233}{1117}.\frac{6713}{1119}-\frac{5}{1119}\)
\(=\frac{1}{1119}\left(\frac{3352}{1117}.4-\frac{2233}{1117}.6713-5\right)\)
\(=\frac{1}{1119}\left(\frac{13408}{1117}-\frac{14990129}{1117}-\frac{5585}{1117}\right)\)
\(=\frac{1}{1119}.\frac{-14982306}{1117}\)
\(=\frac{-14982306}{1249923}\)
Đặt \(x=\frac{1}{1117}\) ,\(y=\frac{1}{1119}\)
Khi đó biểu thức có dạng \((3+x).4y-\left(1+1-x\right).\left(5+1-y\right)-5.y\)
\(=12y+4xy-\left(2-x\right)\left(6-y\right)-5y\)
\(=12y+4xy-\left(12-6x-2y+xy\right)-5y\)
\(=9y+3xy-12+6x\)
\(=\frac{9}{1119}+\frac{3}{1117.1119}-12+\frac{6}{1117}\)
\(=\frac{9.1117+3-12.1117.1119+6.1119}{1117.1119}\)
Ưm đến đây tự tính nha =)