Tính Nhanh:
D = \(\frac{4}{3\cdot7}-\frac{4}{7\cdot11}+\frac{4}{11\cdot15}-\frac{4}{15\cdot19}+\frac{4}{19\cdot23}-\frac{4}{23\cdot27}\)
tính nhanh :
\(B=\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+\frac{1}{11\cdot15}+\frac{1}{15\cdot19}+\frac{1}{19\cdot23}+\frac{1}{23\cdot27}+\frac{1}{27\cdot31}+\frac{1}{31\cdot35}\)
\(A=\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
Phần 1)Đầu tiên bạn nhân B với 1 phần 4 rồi tính đến đoạn gần cuối sẽ ra 1/3 - 1/35 rồi quy đòng rồi tính sẽ ra kêt quả cuối là 32/105 nha
Mình lười lắm nên chỉ help 1 phần thui nha sr
a)\(\frac{1}{2}\)-\(\frac{1}{3\cdot7}\)-\(\frac{1}{7\cdot11}\)-\(\frac{1}{11\cdot15}\)-\(\frac{1}{15\cdot19}\)-\(\frac{1}{19\cdot23}\)-\(\frac{1}{23\cdot27}\)
b) 1-\(\frac{1}{5\cdot10}\)-\(\frac{1}{10\cdot15}\)-\(\frac{1}{15\cdot20}\)-...-\(\frac{1}{95\cdot100}\)
a) \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{19}-\frac{1}{19}+\frac{1}{23}-\frac{1}{23}+\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)
\(=\frac{1}{2}-\frac{2}{27}\)
\(=\frac{23}{54}\)
b) \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+\frac{1}{20}-...-\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
Tính nhanh biểu thức sau :
\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+....+\frac{4}{15\cdot17}+\frac{4}{17\cdot19}+\frac{4}{19\cdot21}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
=\(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
=\(2\left(1-\frac{1}{21}\right)\)
=\(\frac{2.20}{21}=\frac{40}{21}\)
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
tính nhanh và ko quy đồng
a] C = \(\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
b] B = \(\frac{6}{3\cdot5}+\frac{6}{5\cdot7}+\frac{6}{7\cdot9}+.....+\frac{6}{97\cdot99}\)
a) \(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=7.\frac{13}{28}=\frac{7.13}{28}=\frac{13}{4}\)
b) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)
tính E=\(\frac{5}{1\cdot2}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
E=\(\frac{5}{1.2}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
E.\(\frac{1}{7}\)=\(\frac{5}{1.2.7}+\frac{4}{1.11.7}+\frac{3}{11.2.7}+\frac{1}{2.15.7}+\frac{13}{15.4.7}\)
E.\(\frac{1}{7}\)=\(\frac{5}{7.2}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
E.\(\frac{1}{7}\)=\(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
E.\(\frac{1}{7}\)=\(\frac{1}{2}-\frac{1}{28}\)
E.\(\frac{1}{7}=\frac{13}{28}\)
E=\(\frac{13}{28}:\frac{1}{7}=\frac{13}{4}\)
\(E=\frac{5}{1.2}+\frac{1}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(E=\frac{5}{2}+\frac{1}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\)
\(E=\frac{5}{2}+\left(\frac{1}{11}+\frac{3}{22}\right)+\left(\frac{1}{30}+\frac{13}{60}\right)\)
\(E=\frac{5}{2}+\left(\frac{2}{22}+\frac{3}{22}\right)+\left(\frac{2}{60}+\frac{13}{60}\right)\)
\(E=\frac{5}{2}+\frac{5}{22}+\frac{15}{60}\)
\(E=\frac{55}{22}+\frac{5}{22}+\frac{1}{4}\)
\(E=\frac{60}{22}+\frac{1}{4}\)
\(E=\frac{30}{11}+\frac{1}{4}\)
\(E=\frac{120}{44}+\frac{11}{44}\)\(=\frac{131}{44}\)
k mình nha chúc bạn học giỏi
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
giúp mk nha các bn
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{4}{7\cdot11}-\frac{4}{11\cdot15}+...+\frac{4}{91\cdot95}\)
tính
\(\text{So sánh: A=\frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+\frac{1}{8\cdot9}+\frac{1}{10\cdot11}+\frac{1}{12\cdot13}+\frac{1}{14\cdot15}+\frac{1}{16\cdot17}+\frac{1}{18\cdot19} và B=\frac{9}{19}}\)So sánh: A=1/2*3 + 1/4*5 + 1/6*7 + 1/8*9 + 1/10*11 + 1/12*13 + 1/14*15 + 1/16*17 + 1/18*19 và B=9/19
Giúp tớ với, tớ cần gấp !! Cảm ơn nhìu ạ !!
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A