\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{249}\)
Tính nhanh
\(1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
=22/15- 1/1.3 - 1/3.5 - 1/5.7 -.........- 1/11.13 - 1/13.15
=22/15 - (1/1.3+1/3.5+....+1/13.17)
=22/15 - 1/2(2/1.3+2/3.5.........+2/13.17)
=22/15 - 1/2(1-1/3+1/3-1/4+.............+1/13-1/17)
=22/15 - 1/2(1-1/17)
=22/15-8/17
=254/255
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
bạn có thể trình bày cách làm cho mình ko
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\frac{12}{13}\)
\(=\frac{6}{13}\)
1/3 + 1/3×5 + 1/5×7 + 1/7×9 + 1/9×11 + 1/11×13
=> 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +1/9 -1/11 + 1/11 + -1/13
=> 2/3 - 1/13 = 23/39
Đúng 100% nha. k mk nha
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(2A=1-\frac{1}{13}=\frac{12}{13}\)
\(\Rightarrow A=\frac{12}{13}:2=\frac{6}{13}\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)=?????
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
Tính nhanh :
\(C=\frac{1}{3}+\frac{-3}{4}+\frac{3}{5}+\frac{1}{57}+\frac{-1}{36}+\frac{1}{15}+\frac{-2}{9}\)
\(D=\frac{1}{2}+\frac{-1}{5}+\frac{-5}{7}+\frac{1}{6}+\frac{-3}{35}+\frac{1}{3}+\frac{1}{41}\)
\(E=\frac{-1}{2}+\frac{3}{5}+\frac{-1}{9}+\frac{1}{127}+\frac{-7}{18}+\frac{4}{35}+\frac{2}{7}\)
C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9
C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9
C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9
C = 20/57 + -28/36 + 10/15 + -2/9
C = 20/57 + -7/9 + 2/3 + -2/9
C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )
C = 58/57 + -1
C = 1/57
D = 1/2 + -1/5 + -5/7 + 1/6 + -3/35 + 1/3 + 1/41
D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41
D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41
D = 1 + -1 + 1/41
D = 1/41
E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7
E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7
E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7
E = -1 + 5/7 + 1/257 + 2/7
E = -1 + ( 5/7 + 2/7 ) + 1/127
E = -1 + 1 + 1/127
E = 1/127
\(C=\frac{1}{3}+\frac{-3}{4}+\frac{3}{5}+\frac{1}{57}+\frac{-1}{36}+\frac{1}{15}+\frac{-2}{9}.\)
\(C=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}+\frac{2}{9}\right)+\frac{1}{57}\)
\(C=1-1+\frac{1}{57}\)
\(C=\frac{1}{57}\)
D=1/2+-1/5+-5/7+1/6+-3/35+1/3+1/41
D=(1/2+1/6+1/3)-(1/5+3/35+5/7)+1/41
D=1-1+1/41
D=1/41
E=-1/2+3/5+-1/9+1/127+-7/18+4/35+2/7
E=(3/5+4/35+2/7)-(1/2+1/9+7/18)+1/127
E=1-1+1/127
E=1/127
1) \(79\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)
2)\(\left(3\frac{1}{5}-\frac{1}{11}\right).11\)
3)\(4\frac{1}{5}:\left(2\frac{1}{3}.1\frac{2}{5}\right)\)
4)\(13,2.\frac{15}{64}-\left(\frac{4}{5}+\frac{2}{3}\right):3\frac{2}{3}\)
Tính tổng
\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)
\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)
\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)
\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{9.11}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(S=\frac{5}{11}\)
\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)
\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(Q=\frac{1007}{2015}\)
~ Đấng Ed :) ~
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=?\)
Ta có :
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(=\)\(\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{2499}\right)\) ( bước này hơi khó hiểu tí nhé )
\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\) ( phân tích mẫu )
\(=\)\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\) ( áp dụng công thức thoi )
\(=\)\(\frac{1}{2}\left(1-\frac{1}{51}\right)\) ( loại bỏ nhưng phân số đối nhau )
\(=\)\(\frac{1}{2}.\frac{50}{51}\)
\(=\)\(\frac{25}{51}\)
Chúc bạn học tốt ~
ĐẶT \(A\)\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)
\(2.A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(2.A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2.A=1-\frac{1}{51}\)
\(2.A=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}\div2=\frac{25}{51}\)
Tính nhanh nếu có thể:
a)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
b)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)
a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{14}{30}=\frac{7}{15}\)
a)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=2\left(1-\frac{1}{15}\right)\)
\(=2.\frac{14}{15}\)
\(=\frac{28}{15}\)
b)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)
\(...\)