3/3.5+3/5.7+3/7.9+...+3/49.51 = ?
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NHẦM GIẢI LẠI :
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.51}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
=\(\frac{1}{2}.\frac{16}{51}\)
=\(\frac{8}{51}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(=\frac{2}{3}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{51}\right)\)
\(=\frac{2}{3}.\frac{50}{51}=\frac{20}{51}\)
Ủng hộ mk nha !!! ^_^
3/1.3 + 3/3.5 + 3/5.7 + ....... + 3/49.51
= 3 x ( 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/49.51 )
= 3 x ( 1 - 1/51 )
= 3 x 50/51
= 150/151
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
3.2/1.3.2+3.2/3.5.2+3.2/5.7.2+...+3.2/49.51
3/2(2/1.3+2/3.5+2/5.7+....+2/49.51)
3/2(1-1/3+1/3-1/5+1/5-1/7+....+1/49-1/51)
3/2(1-1/51)
3/2 . 50/51
25/17
áp dụng công thức nếu có thừa số thứ 2 ở mẫu trừ đi thừa số thứ 1 bằng số trên tử thi \(\frac{1}{a}-\frac{1}{b}\) ab ở đây là 2 thừa số ở mẫu
VD;3/1.3+3/3.5+...+3/49.51(vì tất cả mẫu trừ cho nhau đều =tử)
nên = 1/1-1/3+1/3+1/5+...+1/49-1/51
=1-1/51
=50/51
a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)
b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(=\frac{15}{16}\)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!1111
\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{49.51}\)
= 3. \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{49}-\frac{1}{51}\right)\)
=\(\frac{3}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
=\(\frac{3}{2}.\frac{16}{51}\)
=\(\frac{8}{17}\)
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