1-2+3-4+5-6+899
1*2*3*4*5*6*7*3*5*7*3*5*3*0*5*6*4*6*3*12*4*76*7*899*45=?
1.1+2=1+2
2.3^2=18
3.6+3=21
4.5+6+3+4+8+97+6=144
5.01+65+559+555+899=2083
a ) 1 + 2 = 1 + 2
b ) 3 ^ 2 = 9
c ) 6 + 3 = 9
d ) 5 + 6 + 3 + 4 + 8 + 97 + 6 = 129
đ ) 01 + 65 + 559 + 555 + 899 = 2079
Tick mình đi mình tick lại cho !
Câu 1 : (11/4.-5/9-4/9.11/4).8/33
Câu 2 : (17/28+18/29-19/30-20/31).(-5/2+1/4+1/6)
Câu 3 :(1/2+1).(1/3+1).(1/4+1) ... (1/99+1)
Câu 4 : 3/2^2.8/3^2.15/4^2 ... 899/30^2
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
Câu 2: (\(\dfrac{17}{28}\) + \(\dfrac{18}{29}\) - \(\dfrac{19}{30}\) - \(\dfrac{20}{31}\)).(-\(\dfrac{5}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{6}\))
= (\(\dfrac{17}{28}\) + \(\dfrac{18}{29}\) - \(\dfrac{19}{30}\) - \(\dfrac{20}{31}\)).(-\(\dfrac{5}{2}\) + \(\dfrac{5}{12}\))
= (\(\dfrac{17}{28}\) + \(\dfrac{18}{29}\) - \(\dfrac{19}{30}\) - \(\dfrac{20}{31}\)). 0
= 0
a) A=3/4*8/9*15/16+...+899/900 b)B=1/1*2*3+1/2*3*1+1/3*4*5+...+1/98*99*100
c)C=1/2+1/14+1/35+1/65+1/104+1/152 d) D=1/1*2*3*4+1/2*3*4*5+1/3*4*5*6+...+1/27*28*29*30
giải giúp mk
a,
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)
b,
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)
c, (Chịu :V)
d,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
Chúc bạn học tốt nha.
a. 1! + 2 . 2! + 3 . 3! + ... + 10 . 10!
b. 1 . 2 + 2. 5 + 3 . 8 + ... + 50 . 149
c. 1 . 4 + 2 . 5 + 3 . 6 + .. + 50 . 53
d. \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
Các bạn giúp mk nhóa!
a) A=3/4*8/9*15/16...899/900 b)B=1/1*2*3+1/2*3*1+1/3*4*5+...+1/98*99*100 c)C=1/2+1/14+1/35+1/65+1/104+1/152 d) D=1/1*2*3*4+1/2*3*4*5+1/3*4*5*6+...+1/27*28*29*30
giair giups mk
Tìm y:
-y:\(\dfrac{1}{2}\)-\(\dfrac{5}{2}\)=4\(\dfrac{1}{2}\)
Tính:
N = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)....\(\dfrac{899}{900}\).\(\dfrac{960}{961}\)
S=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{10.11.12}\)+\(\dfrac{1}{11.12.13}\)
Tìm y:
-y:1/2-5/2=4+1/2
-y:1/2 = 4+1/2+5/2
-y:1/2 = 7
-y = 7.2
y = -14
Vậy y = -14
Tính:
a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}_{ }\)
b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
c, \(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)
Bài giải
a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)
\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)
b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)
1. tìm tích của A= \(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times..\times\frac{899}{900}\)
2. CMR \(\frac{1}{5}+\frac{1}{6}+..+\frac{1}{17}< 2\)
3. tính \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+..+\frac{1}{10.11.12}\)
3. \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\)
\(\Leftrightarrow2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{10.11.12}\)
\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{11.12}\)
\(\Leftrightarrow2M=\frac{1}{2}-\frac{1}{132}\)
\(\Leftrightarrow2M=\frac{65}{132}\)
\(\Leftrightarrow M=\frac{65}{132}\div2\)
\(\Leftrightarrow M=\frac{65}{264}\)
1\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)
\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)
\(\Leftrightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)
\(\Leftrightarrow A=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4...30\right)\left(2.3.4...30\right)}\)
\(\Leftrightarrow A=\frac{1.31}{30.2}\)
\(\Leftrightarrow A=\frac{31}{60}\)
2. Đặt \(A=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)
\(\Rightarrow A< \frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\)
\(\Rightarrow A< 1+1=2\)
Vậy a < 2 (đpcm)