Tính A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{99.101}\right)\)
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Tính nhanh: \(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
Tính \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{99.101}\right)\)
4/3 .9/8 .16/15 ......10000/9999
2.2 .3.3.4.4.....100.100 /1.3.2.4.3.5.....99.101
( 2.3.4 ....100 ) .( 2.3.4 ....100) / ( 1.2.3.....99). (3.4.5...101 )
100*2 /101
200/101
chú thích không có trong bài nhé
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" ..........." là dấu nhân
" / " là dâu của phân số
" * " cũng là dấu nhân nha bạn
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\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
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\(B=\left(\frac{1}{1.3}+1\right).\left(\frac{1}{2.4}+1\right).\left(\frac{1}{3.5}+1\right).....\left(\frac{1}{99.101}+1\right)\)
tính nhanh
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{99.101}\right)\)
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{99.101}\right)\)
Ta có : A = (4/1.3) . (9/2.4).......(10000/99.101)
= (2.2/1.3). (3.3/2.4).......(100.100/99.101)
=(2.3.4......99.100/1.2.3.....98.99 ) . ( 2.3.4.......100/3.4.5.....101)
=(100/1) . ( 2/101 )
=200/101
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Tính:
\(A=\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
\(B=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
Mong các bạn giúp đỡ. Cảm ơn nhiều *cúi đầu*
\(A=xemlai\) chưa hưa hiểu Quy luật
\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)
\(B=\frac{2.100}{1.101}=\frac{200}{101}\)
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Cho biểu thức: C = \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\). Chứng minh rằng C không phải là số nguyên
Tính các tích sau:
a)\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
b)\(B=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)......\left(1-\frac{1}{1326}\right)\)
c)\(C=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{99.101}\right)\)
Tính nhanh:
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right)....\left(1-\frac{1}{10000}\right)\)
\(B=\left(\frac{1}{1.3}+1\right).\left(\frac{1}{2.4}+1\right).\left(\frac{1}{3.5}+1\right)....\left(\frac{1}{99.101}+1\right)\)
Làm câu nào cx được,nhanh nhé