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A=\(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
\(A=\frac{1}{2.9}+\frac{1}{9.7}+...+\frac{1}{252.509}\)
\(A=\frac{2}{4.9}+\frac{2}{9.14}+...+\frac{2}{504.509}\)
\(A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{504.509}\right)\)
\(A=\frac{2}{5}.\left(\frac{9-4}{4.9}+\frac{14.9}{9.14}+...+\frac{509-504}{504.509}\right)\)
\(A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}.\frac{505}{2036}\)
\(A=\frac{101}{1018}\)
\(B=\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(\frac{1}{2}B=\frac{1}{10.9.2}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405.2}\)
\(\frac{1}{2}B=\frac{1}{10.18}+\frac{1}{18.26}+\frac{1}{26.34}+...+\frac{1}{802.810}\)
\(4B=\frac{8}{10.18}+\frac{8}{18.26}+\frac{8}{26.34}+...+\frac{8}{802.810}\)
\(4B=\frac{18-10}{10.18}+\frac{26-18}{28.26}+\frac{34-26}{26.34}+...+\frac{810-802}{802.810}\)
\(4B=\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\)
\(4B=\frac{1}{10}-\frac{1}{810}\)
\(4B=\frac{8}{81}\)
\(B=\frac{2}{81}\)
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TẬP HỢP RA HAI NHÓM .MỘT NHÓM SỐ ÂM.CÒN NHÓM KIA LÀ SỐ DƯƠNG MÀ TÍNH
STUDY WELL
K NHA
MK XIN CẢM ƠN CÁC BẠN NHÌU
C = 24.7 −35.9 +27.10 −39.13 +...+2301.304 −3401.405
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
Study well
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
Đặt Biểu thức trên là A
\(A=\frac{1}{2.5.9}+\frac{1}{2.9.13}+\frac{1}{2.13.17}+...+\frac{1}{2.397.401}+\frac{1}{2.401.405}\)
\(A=\frac{1}{2}\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{397.401}+\frac{1}{401.405}\right)\)
\(4A=\frac{1}{2}\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{397.401}+\frac{4}{401.405}\right)\)
\(4A=\frac{1}{2}\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{401-397}{397.401}+\frac{405-401}{401.405}\right)\)
\(4A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{405}\right)=\frac{1}{2}.\frac{80}{405}=\frac{40}{405}\Rightarrow A=\frac{40}{4.405}=\frac{2}{81}\)
a)b) Bạn nhân cả tử và mẫu với 2. Mình làm luôn, ko ghi lại đề bài
a)\(\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
=\(\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
=\(\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b)\(\frac{2}{10.18}+\frac{2}{18.26}+\frac{2}{26.34}+...+\frac{2}{802.810}\)
=\(\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\right)\)
=\(\frac{1}{4}.\left(\frac{1}{10}-\frac{1}{810}\right)\)
=\(\frac{1}{4}.\frac{8}{81}=\frac{2}{81}\)
c) Mình biết làm, ddoiwtj tí nữa mình làm cho. Giờ đang mỏi tay
Thẳng Nobita kun có chép bài thì đừng t..i..c..k cho nó
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