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Đặt \(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+\frac{7^2}{23.30}\)
\(\Rightarrow A=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
\(\Rightarrow A=7.\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(\Rightarrow A=\frac{49}{15}\)
đặt biểu thức là B
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có :
\(B=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{23}-\frac{1}{30}\right)\)
\(B=7.\left(\frac{1}{2}-\frac{1}{30}\right)=7.\frac{7}{15}=\frac{49}{15}\)
Ai thấy đúng thì ủng hộ nha !!!
A = 7 (7 / 2.9 + 7 / 9.16 + .......... + 7/65.72)
A=7( 1/2 - 1/9 +1/9 - 1/16 +......+1/65 - 1/72)
A= 7 ( 1/2 -1/72)
A= 7 . 35/72
A=245/72
\(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7}{16.23}+.....+\frac{7^2}{65.72}\)
=\(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)
=\(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
=\(7.\frac{35}{72}\)
=\(\frac{245}{72}\)
Ta có:
C = \(\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)
=> C = \(7.\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+...+\frac{7}{65.72}\right)\)
=> C = \(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)
=> C = \(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
=> C = \(7.\frac{35}{72}=\frac{245}{72}\)
Nhìn kĩ là ra thôi :
\(\frac{7^2}{2.9}+\frac{7^2}{9.16}+...+\frac{7^2}{65.72}\)
= \(7\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right)\)
= \(7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right)\)
= \(7\left(\frac{1}{2}-\frac{1}{72}\right)\)
= \(7.\frac{35}{72}=3\frac{29}{72}\)
Ta có:\(\frac{1}{2.9}=\frac{1}{2}-\frac{1}{9}\)
\(\frac{1}{9.7}=\frac{1}{9}-\frac{1}{7}\)
\(⋮\)
\(\frac{1}{252.504}=\frac{1}{252}-\frac{1}{504}\)
\(A=\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-...............+\frac{1}{252}-\frac{1}{504}\)
\(A=\frac{1}{2}-\frac{1}{504}\)
\(A=\frac{251}{504}\)
Đặt \(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)
\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(\Leftrightarrow A=\frac{2}{5}.\frac{505}{2036}\)
\(\Leftrightarrow A=\frac{101}{1018}\)
~ Hok tốt ~
#)Giải :
\(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)
\(A=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}\times\frac{505}{2036}\)
\(A=\frac{101}{1018}\)
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
Hướng dẫn:
\(7B=\dfrac{9-2}{2.9}+\dfrac{16-9}{9.16}+\dfrac{23-16}{16.23}+...+\dfrac{107-100}{100.107}+\dfrac{114-107}{107.114}\)
\(=\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{100}-\dfrac{1}{107}+\dfrac{1}{107}-\dfrac{1}{114}\)
\(=\dfrac{1}{2}-\dfrac{1}{114}=...\)
\(B=\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+...+\frac{1}{100.107}+\frac{1}{107.114}\)
\(B=\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{100}-\frac{1}{107}+\frac{1}{107}-\frac{1}{114}\right)\)
\(B=\frac{1}{7}\left(\frac{1}{2}-\frac{1}{114}\right)\)
\(B=\frac{1}{7}\left(\frac{57}{114}-\frac{1}{114}\right)\)
\(B=\frac{1}{7}.\frac{56}{114}\)
\(B=\frac{1}{7}.\frac{28}{57}\)
\(B=\frac{4}{57}\)