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a) \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}+\frac{3}{418}+\frac{3}{550}\)
= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+\frac{3}{19.22}+\frac{3}{22.25}\)
= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
= \(\frac{1}{1}-\frac{1}{25}\)
= \(\frac{24}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)
= \(\frac{1}{1}-\frac{1}{2n+3}\)
= \(\frac{2n+2}{2n+3}\)
c) \(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{7070707}{15151515}\)
= \(\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{7.1010101}{15.1010101}\)
= \(\frac{7}{15}-\frac{7}{15}\)
= 0
Ta có :
\(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{70707070}{15151515}\)
\(=\)\(\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{14}{3}\)
\(=\)\(\frac{7}{15}-\frac{14}{3}\)
\(=\)\(\frac{-21}{5}\)
Chúc bạn học tốt ~ ( cái chỗ \(\frac{70770707}{15151515}\) nếu có nhầm thì bạn sử giùm mk nhé )
Đặt \(A=\frac{15+\frac{15}{7}-\frac{15}{11}+\frac{15}{2009}-\frac{15}{13}}{\frac{4}{2009}-\frac{4}{13}+\frac{4}{7}-\frac{4}{11}+4}\)
\(=\frac{15\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}{4\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}\)
\(=\frac{15}{4}\)
Đặt \(B=\frac{5\cdot2010-1996}{14+4\cdot2010}\)
\(=\frac{5\left(1996+4\right)-1996}{14+4\cdot2010}\)
\(=\frac{5\cdot1996+20-1996}{14+4\left(1996+4\right)}\)
\(=\frac{4\cdot1996+20}{4\cdot1996+30}\)
\(\Rightarrow A\cdot B=\frac{4\cdot1996+20}{4\cdot1996+30}\cdot\frac{15}{4}=\frac{15\cdot4\left(1996+5\right)}{4\left(4\cdot1996+30\right)}=\frac{15\left(1996+5\right)}{4\cdot1996+30}=\frac{30015}{8004}\)
mặc dầu ko khoa học lắm nhưng mình thấy cũng được đấy
(\(\frac{15}{22}\)-\(\frac{7}{13}\)) x \(\frac{13}{7}\)- \(\frac{15}{22}\) x(\(\frac{6}{7}\)- \(\frac{22}{15}\))
=(\(\frac{15}{22}\)x \(\frac{13}{7}\)- 1) - \(\frac{15}{22}\)x\(\frac{6}{7}\)-1
=\(\frac{15}{22}\)x \(\frac{13}{7}\)-\(\frac{15}{22}\)x\(\frac{6}{7}\)(vì -1+1=0)
=\(\frac{15}{22}\)x(\(\frac{13}{7}\)-\(\frac{6}{7}\))
=\(\frac{15}{22}\)
làm hơi tắt nha
\(\left(\frac{15}{22}-\frac{7}{13}\right).\frac{13}{7}-\frac{15}{22}.\left(\frac{6}{7}-\frac{22}{15}\right)\)
\(=\frac{15}{22}.\frac{13}{7}-\frac{7}{13}.\frac{13}{7}-\frac{15}{22}.\frac{6}{7}+\frac{15}{22}.\frac{22}{15}\)
\(=\frac{15}{22}.\left(\frac{13}{7}-\frac{6}{7}\right)-1+1\)
\(=\frac{15}{22}.1-1+1\)
\(=\frac{15}{22}-\left(1-1\right)\)
\(=\frac{15}{22}-0\)
\(=\frac{15}{22}\)
7/13(7/15+8/15)-(5/12).(7/13)
=7/13(1-5/12)=7/13(7/12)=49/(12.13)
7/13 . 7/15 - 5/12 . 7/13 + 7/13 . 8/15
=49/195 - 35/156 + 56/195
=49/156
34 −35 +37 +311 134 −135 +137 +1311
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{13.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\)
\(=\frac{3}{13}\)
Ta có: \(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)
\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)
\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{8}{99}\)
\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)
\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\frac{32}{99}\)
\(A=\frac{8}{99}\)
hình như em đánh thừa số 1 rồi
\(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{707070}{151515}\)
\(=\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{93}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{707070}{151515}\)
\(=\frac{7}{15}-\frac{707070}{151515}=\frac{7}{15}-\frac{7}{15}=0\)
(do \(\frac{707070}{151515}=\frac{7.101010}{15.101010}=\frac{7}{15}\))