3 1/11 * 27 27/46 * 1 6/ 17 * 2 4/9
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
NL
2
LD
1
NT
1
NT
5
VD
11 tháng 12 2016
\(3\frac{3}{11}x27\frac{27}{46}x1\frac{6}{17}x2\frac{4}{9}=\frac{34x1269x23x22}{11x46x17x9}=\frac{2x17x141x9x23x2x11}{11x23x17x9}=282\)
KS
0
TH
1
19 tháng 7 2019
\(25\frac{2}{13}-\left(\frac{15}{17}+15\frac{2}{3}\right)=\frac{327}{13}-\left(\frac{15}{17}+\frac{47}{3}\right)\)
\(=\frac{327}{13}-\frac{844}{51}\)
\(=\frac{5705}{663}\)
\(\frac{5}{30}+\frac{15}{90}+\frac{250}{150}+\frac{350}{210}+\frac{45}{270}=\frac{1}{6}+\frac{1}{6}+\frac{5}{3}+\frac{5}{3}+\frac{1}{6}\)
\(=\frac{3}{6}+\frac{10}{3}\)
\(=\frac{1}{2}+\frac{10}{3}\)
\(=\frac{23}{6}\)
\(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{17}.2\frac{4}{9}=\frac{34}{11}.\frac{27}{46}.\frac{23}{17}.\frac{22}{9}\)
\(=\frac{68}{9}.\frac{27}{34}\)
\(=6\)
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
Cho mk hỏi bạn có chép sai đầu bài ko