\(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-\frac{37}{27.64}-...-\frac{331}{1000.1331}\)
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a) Ta có: \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}=\frac{4.\left(\frac{1}{7}-\frac{1}{177}-\frac{1}{1779}\right)}{5.\left(\frac{1}{7}-\frac{1}{177}-\frac{1}{1779}\right)}=\frac{4}{5}\)
b) \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-.....-\frac{331}{1000.1331}\)
\(=\frac{1330}{1331}-\left(\frac{8-7}{1.8}+\frac{27-8}{8.27}+.....+\frac{1331-1000}{1000.1331}\right)\)
\(=\frac{1330}{1331}-\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{27}+....+\frac{1}{1000}-\frac{1}{1331}\right)\)
\(=\frac{1330}{1331}-\left(1-\frac{1}{1331}\right)\)
\(=\frac{1330}{1331}-\frac{1330}{1331}=0\)
Vậy \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-....\frac{331}{1000.1331}=0\)
CHÚC BẠN HỌC TỐT
a) \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}\)
\(=\frac{4\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}{5\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}\)
\(=\frac{4}{5}\)
\(\frac{m}{n}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{1331}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1330}\right)\)
\(\frac{m}{n}=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1330}+\frac{1}{1331}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1330}\right)\)
\(\frac{m}{n}=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1330}+\frac{1}{1331}\right)-\left(1+\frac{1}{2}+...+\frac{1}{665}\right)\)
\(\frac{m}{n}=\frac{1}{666}+\frac{1}{667}+...+\frac{1}{1330}+\frac{1}{1331}\)
\(\frac{m}{n}=\left(\frac{1}{666}+\frac{1}{1331}\right)+\left(\frac{1}{667}+\frac{1}{1330}\right)+...+\left(\frac{1}{998}+\frac{1}{999}\right)\)
\(\frac{m}{n}=\frac{1997}{666.1331}+\frac{1997}{667.1330}+...+\frac{1997}{998.999}=\frac{1997k_1+1997.k_2+...+1997.k_{333}}{666.667...1331}\)
\(\frac{m}{n}=\frac{1997.\left(k_1+k_2+...+k_{333}\right)}{666.667...1330.1331}\) trong đó: k1;...; k333 là các thừa số phụ của các phân số trong tổng
Nhận xét: phân số trên có tử chia hết cho 1997 là số nguyên tố; mẫu số không chia hết cho thừa số nguyên tố 1997 nên khi rút gọn tử vẫn chia hết cho 1997
=> m chia hết cho 1997
\(\frac{7}{13}.\frac{5}{19}+\frac{7}{19}.\frac{8}{13}-\frac{37}{19}=\frac{1}{19}\left(\frac{5.7}{13}+\frac{7.8}{13}-37\right)=\frac{1}{19}\left(\frac{91}{13}-37\right)=\frac{1}{19}.\left(-30\right)=-\frac{30}{19}\)
\(\left(\frac{2}{9}+\frac{7}{9}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)=1+1=2\)\(2\)
\(\frac{19}{37}+1-\frac{19}{37}=\left(\frac{19}{37}-\frac{19}{37}\right)+1=1\)
\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}\right):\left(\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}.\left[\frac{12\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right]:\left[\frac{4\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right].\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}\left(\frac{12}{3}.\frac{5}{4}\right).\frac{124242423}{237373735}\)
\(B=1\frac{6}{41}.5.\frac{123}{235}\)
\(B=\frac{47.5.123}{41.235}=\frac{47.5.41.3}{41.5.47}=3\)
B=\(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)
B=\(\frac{47}{41}.\left(\frac{12.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}:\frac{4.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right).\frac{123.1010101}{235.1010101}\)
B=\(\frac{47}{41}.\left(\frac{12}{3}:\frac{4}{5}\right).\frac{123}{235}=\frac{47}{41}.\left(\frac{12}{3}.\frac{5}{4}\right).\frac{123}{235}\)
B=\(\frac{47}{41}.\frac{15}{3}.\frac{123}{235}=\frac{47.5.3.41.3}{41.3.5.47}=3\)
Vậy B=3
Chúc bn học tốt
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
nguyễnCôngBắcKỳ_6a1, bài này còn cần nữa k, mik lm cho