tínhA = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)......\left(1+\frac{1}{100}\right)\)
Tính
a) \(\left( {\frac{4}{5} - 1} \right):\frac{3}{5} - \frac{2}{3}.0,5\)
b) \(1 - {\left( {\frac{5}{9} - \frac{2}{3}} \right)^2}:\frac{4}{{27}}\)
c)\(\left[ {\left( {\frac{3}{8} - \frac{5}{{12}}} \right).6 + \frac{1}{3}} \right].4\)
d) \(0,8:\left\{ {0,2 - 7.\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
A=\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)....\left(\frac{1}{99}+1\right)\)
b) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)..........\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)
\(=-\frac{1}{2}.-\frac{2}{3}..........-\frac{99}{100}\)
\(=\frac{-1}{100}\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5.....100}{2.3.4.....99}\)
\(=\frac{100}{2}=50\)
tính :
a)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)\)
b) B=\(\left(1-\frac{1}{^{2^2}}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{100^2}\right)\)
a/ \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)=\frac{3}{2}\times\frac{4}{3}\times....\times\frac{101}{100}=\frac{101}{2}\)
b/ Tự chép đề nha\(B=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)....\left(1-\frac{1}{100}\right)\left(1+\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{3}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{99}{100}\times\frac{101}{100}=\frac{1}{2}\times\frac{101}{100}=\frac{101}{200}\)
Đề a) (1+1/2) (1+1/3) (1+1/4)...(1+1/100)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{4}{3}....\frac{101}{100}=\frac{3.4...101}{2.3...100}=\frac{101}{2}\)
Học tốt
Thực hiện phép tính :
a, A =\(\left(1:\frac{5^2}{10^2}\right).\left(1\frac{1}{1}\right)^2+25.\left[1:\left(\frac{4}{3}\right)^2:\left(\frac{5}{4}\right)^3\right]:\left(1:\frac{-8}{27}\right)\)
b, B =\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{100^2}\right)\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
Tính nhanh
a, \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
b, \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)\)
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
b) = \(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=-1\left(\frac{1.2.3...99}{2.3.4...100}\right)=-\frac{1}{100}\)
Tính :
\(\left[\frac{1}{100}-1^2\right].\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right].\left[\frac{1}{100}-\left(\frac{1}{3}\right)^2\right]...\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)
Tính:
\(\left[\frac{1}{100}-1^2\right].\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right].\left[\frac{1}{100}-\left(\frac{1}{3}\right)^2\right].....\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)
Xét : \(\frac{1}{100}-\frac{1}{n^2}=\frac{n^2-100}{100n^2}=\frac{\left(n-10\right)\left(n+10\right)}{100n^2}\)
Áp dụng , đặt biểu thức cần tính là A , ta có :
\(A=\left(\frac{1}{100}-\frac{1}{1^2}\right)\left(\frac{1}{100}-\frac{1}{2^2}\right)\left(\frac{1}{100}-\frac{1}{3^2}\right)...\left(\frac{1}{100}-\frac{1}{20^2}\right)\)
\(=\frac{\left(1-10\right)\left(1+10\right)}{100.1^2}.\frac{\left(2-10\right)\left(2+10\right)}{100.2^2}.\frac{\left(3-10\right)\left(3+10\right)}{100.3^2}...\frac{\left(10-10\right)\left(10+10\right)}{100.10^2}...\frac{\left(20-10\right)\left(20+10\right)}{100.20^2}\)
Nhận thấy trong A có một nhân tử (10-10) = 0 nên A = 0
làm thế thì hơi dài đấy Hoàng Lê Bảo Ngọc
ta nhận thấy trong biểu thức chứa thừa số \(\frac{1}{100}-\left(\frac{1}{10}\right)^2=\frac{1}{100}-\frac{1}{100}=0\)
=>biểu thức ấy =0
Nguyễn Thiều Công Thành Ừ , tại mình quên không để ý :)
\(B=\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):...:\left(\frac{1}{98}-1\right):\left(\frac{1}{99}-1\right):\left(\frac{1}{100}-1\right)\)
\(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):....:\left(\frac{1}{100}-1\right)\text{ có số số lẻ thừa số âm nên bằng:}\)
\(-\left[\left(1-\frac{1}{2}\right):\left(1-\frac{1}{3}\right):...\left(1-\frac{1}{100}\right)\right]=-\left[\frac{1}{2}:\frac{2}{3}:\frac{3}{4}:......:\frac{99}{100}\right]=-\left(\frac{1.3.4...100}{2.2.3...99}\right)=-50\)
bn thử nghĩ xem
mk chắc chắn bn lm được]