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biết làm bài 1 thôi
\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)
= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)
lượt bỏ đi còn :
\(\frac{1000}{2}=500\)
Bài 2:
B = (1/2 - 1)(1/3 -1).(1/4 -1)...(1/1000 - 1)
B = (1/2 - 2/2).(1/3 - 3/3)...(1/1000 - 1000/1000)
B = (-1/2).(-2/3)...(-999/1000)
Xét dãy số: 1; 2 ;3;...999
Dãy số trên có 999 số hạng vậy B là tích của 999 số âm
B = - 1/1000
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
Câu a:
A = (1\(\frac16\) x \(\frac67\) x 6 : \(\frac35\)) : (4\(\frac15\) x \(\frac{10}{11}\) + 5\(\frac{2}{10}\))
A = (7/6 x 6/7 x 6 x 5/3) :(21/5 x 10/11 + 52/10)
A = (1 x 6 x 5/3) : (210/55 + 26/5)
A = 10 : (210/55 + 286/55)
A = 10 : (496/55)
A = 10 x 55/496
A = 275/248
Câu b:
B = \(1\)\(\frac{13}{15}\) x 25% x 3 + (8/15 - 79/60) : 1\(\frac{23}{4}\)
B = 28/15 x 1/4 x 3 + (32/60 - 79/60) : 27/4
B = 7/15 x 3 - 47/60 x 4/27
B = 7/5 - 47/405
B = 104/81
a, \(A=\frac{22}{27}\)
b,\(B=\frac{1}{57}\)
C,\(C=\frac{1}{50}\)
d, \(D=0\)
Câu a:
A = -1/2 - (-3)/5 + (-1/9) + 1/27 + 7/18 + 4/35 - (-2/7)
A = -1/2 + 3/5 - 1/9 + 1/27 + 7/18 + 4/35 + 2/7
A = (-1/2 - 1/9 + 7/18 + 1/27) + (3/5 + 4/35 + 2/7)
A = (-27/54 - 6/54 + 21/54 + 2/54) + (21/35 + 4/35 + 10/35)
A = -10/54 + 1
A = -5/27 + 1
A = 22/27
\(B=\frac{12}{11}x\frac{13}{12}x.......x\frac{16}{15}\)
\(=\frac{16}{11}\)
Bài 1a:
A = \(\frac{7^2}{7.8}\times\frac{8^2}{8.9}\times\ldots\times\frac{11^2}{11.12}\)
A = \(\frac{7.8.9.\ldots11}{7.8.\ldots11}\) x \(\frac{7.8.9\ldots11}{8.9.12}\)
A = 7/12
Bài 1a:
A = \(\frac{7^2}{7.8}\times\frac{8^2}{8.9}\times\ldots\times\frac{11^2}{11.12}\)
A = \(\frac{7.8.9.\ldots11}{7.8.\ldots11}\) x \(\frac{7.8.9\ldots11}{8.9.12}\)
A = 7/12
Bài 1B
B = (1+ 1/11).(1 + 1/12)...(1+ 1/15)
B = (11/11 + 1/11).(12/12 + 1/12)...(15/15+ 1/15)
B = 12/11.13/12....16/15
B = 16/11
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
Câu b: Đặt \(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2004}-1\right)\)
Ta có: \(\frac{1}{2}-1=\left(-\frac{1}{2}\right);\frac{1}{3}-1=\left(-\frac{2}{3}\right);...;\frac{1}{2004}-1=\left(-\frac{2003}{2004}\right)\)
\(\Rightarrow B=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{2003}{2004}\right)\)
Vì B là 2003 thừa số âm nhân lại với nhau nên B là số âm
\(\Rightarrow B=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\right)=-\frac{1}{2004}\)
Câu a: Đặt \(A=1+2^4+2^8;B=1+2+2^2+...+2^{11}\)
\(\Rightarrow16A=2^4+2^8+2^{12}\) \(\Rightarrow15A=2^{12}-1\) \(\Rightarrow A=\frac{2^{12}-1}{15}\) \(\left(1\right)\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{12}\) \(\Rightarrow B=2^{12}-1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)