B=1-1212(1+2)-1313(1+2+3)-1414(1+2+3+4)-...-120120(1+2+3+...+20)
Rút gọn các phân số sau:
a) − 1212 − 2424 ;
b) 120120 − 240240 ;
c) 1313 − 1414 .
a ) − 1212 − 2424 = ( − 1212 ) : ( − 1212 ) ( − 2424 ) : ( − 1212 ) = 1 2
b ) 120120 − 240240 = 120120 : ( − 120120 ) − 240240 : ( − 120120 ) = − 1 2
c) 1313 − 1414 = 1313 : ( − 101 ) − 1414 : ( − 101 ) = − 13 14
Rút gọn các phân số sau
a ) 33 66 ; b ) − 22 77 ; c ) 3030 6060
d ) − 1212 − 2424 ; e ) 120120 − 240240 ; f ) 1313 − 1414
a) 33 66 = 33 : 33 66 : 33 = 1 2 b ) − 22 77 = − 22 : 11 77 : 11 = − 2 7
c ) 3030 6060 = 3030 : 3030 6060 : 3030 = − 1 2 d ) − 1212 − 2424 = ( − 1212 ) : ( − 1212 ) ( − 2424 ) : ( − 1212 ) = 1 2
e ) 120120 − 240240 = 120120 : ( − 120120 ) − 240240 : ( − 120120 ) = − 1 2
f ) 1313 − 1414 = 1313 : ( − 101 ) − 1414 : ( − 101 ) = − 13 14
tìm B B = (1 - 12) x (1 - 13) x (1 - 14 x (1 - 15) x .... x (1 - 12003) x (1 - 12004)
Lúc nãy, cô còn dạy học nên giờ cô mới giảng cho em được nhé.
B = (1 - \(\dfrac{1}{2}\))\(\times\)(1 - \(\dfrac{1}{3}\))\(\times\)(1 - \(\dfrac{1}{4}\))\(\times\)(1-\(\dfrac{1}{5}\))\(\times\)...\(\times\)(1- \(\dfrac{1}{2003}\))\(\times\)(1-\(\dfrac{1}{2004}\))
B = \(\dfrac{2-1}{2}\)\(\times\)\(\dfrac{3-1}{3}\)\(\times\)\(\dfrac{4-1}{4}\)\(\times\)\(\dfrac{5-1}{5}\)\(\times\)...\(\times\)(\(\dfrac{2003-1}{2003}\))\(\times\)(\(\dfrac{2004-1}{2004}\))
B = \(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\)\(\times\)\(\dfrac{3}{4}\)\(\times\)\(\dfrac{4}{5}\)\(\times\)...\(\times\)\(\dfrac{2002}{2003}\)\(\times\)\(\dfrac{2003}{2004}\)
B = \(\dfrac{2\times3\times4\times...\times2003}{2\times3\times4\times...\times2003}\)\(\times\) \(\dfrac{1}{2004}\)
B = \(\dfrac{1}{2004}\)
Bài 1: Tính nhanh
a) \(\frac{1111+1212+1313+1414+1515+1616}{2020+2121+2222+2323+2424+2525}\)
b) \(\frac{5,4:0,4\cdot1420+4,5\cdot780\cdot3}{3+6+9+12+......+24+27}\)
c) \(\frac{7,2:2\cdot28,6+1,43\cdot2\cdot64}{2+2+4+6+10+16+......+110}\)
Bài 2: Tính tổng
D = 1 x 2 + 2 x 3 + 3 x 4 + ...... +99 x 100
bài 1: Tính biểu thức 1 cách hợp lý\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)
bài 2: Tím y là số tự nhiên
\(2< \)( \(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\)):5 x y \(< \frac{5}{6}\)
\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)
\(=\frac{101.10+101.11+...+101.17}{101.20+101.21+...+101.27}\)
\(=\frac{101.\left(10+11+...+17\right)}{101.\left(20+21+...+27\right)}\)
\(=\frac{108}{188}\)
\(=\frac{27}{47}\)
\(2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right)\cdot5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{20}{120}+\frac{16}{120}+\frac{9}{120}+\frac{5}{120}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{5}{12}:5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{1}{12}.y>\frac{5}{6}\)
Đặt :\(\frac{1}{12}.y=2\Rightarrow y=2:\frac{1}{12}=24\)
\(\frac{1}{12}.y=\frac{5}{6}\Rightarrow y=\frac{5}{6}:\frac{1}{12}=10\)
\(\Rightarrow24>y>10\)
\(\Rightarrow y\in\left\{11;12;...;23\right\}\)
Tìm x :
1313 x 1414 - 1414 x X = 1414 x 13
( 1 x 2 + 2 x 3 + 3 x 4 + ...+ 50x 51 ) x X = 2 x 4 + 4 x 6 + 6 x 8 + ...+ 100 x 102
X x ( X + 5 ) - 7 x ( X + 5 ) = 0
Bài 5:
a,1111+1212+1313+1414+1515+1616/2020+2121+2222+2323+2424+2525
b,5.4:0,4x1420+4,5x780x3/3+6+9+12+.....+24+27
c,7,2:2x28,6+1,43x2x64/2+2+4+6+10+16+...110
Câu 1 1+2+3+...+97+98+99
Câu 2 1010+1111+1212+1313+...+9898+9999
Câu 1: 1+2+3+...+97+98+99 ( có 99 số hạng )
= ( 99 + 1 ) x 99 : 2
= 4950
Câu 2: 1010+1111+1212+1313+...+9898+9999 có ( 9999 - 1010 ) : 101 + 1 = 90 số
= ( 9999 + 1010 ) x 90 : 2
= 495405
Câu 1 :
Dãy số trên có số số hạng là :
( 99 - 1 ) : 1 + 1 = 99 ( số )
Tổng của dãy số trên là :
( 99 + 1 ) x 99 : 2 = 4950
Đáp số : 4950
Câu 2 :
Dãy trên có số số hạng là :
( 9999 - 1010 ) : 101 + 1 = 90 ( số )
Tổng của dãy số trên là :
( 9999 + 1010 ) x 90 : 2 = 495405
Đáp số : 495405
Bài 1: A=2/3*7 + 2/7*11 + 2/11*15+ ... +2/99*103 Bài 2: A=7/2 + 7/6 + 7/12 + 7/20 + 7/30 + 7/42 + 7/56 + 7/72 + 7/90 Bài 3: A=505/10*1212 + 505/12*1414 + 505/14*1616 +...+ 505/96*9898 Bài 4: A=2/1*3 - 4/3*5 - 6/5*7 - ... - 20/19*21 Bài 5: A=1 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56 - 17/72 + 19/90 :>
\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
Bài 1:
Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)
\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)
Bài 2:
Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)
\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)
\(=7\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{63}{10}\)