Tính \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{99.101}\right)\)
Những câu hỏi liên quan
Tính A=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{99.101}\right)\)
= 4/1.3 x 9/2.4 x 16/3.5 x...x 10000/99.101
= 2.2/1.3 x 3.3/2.4 x 4.4/3.5 x..x 100.100/99.101
= (2.3.4. ... 100/1.2.3. .... 99) x (2.3.4. ... .100/3.4.5. ... .101)
= 100.2/101
=200/101
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\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)
\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)
\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)
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\(A=\left(1+\frac{1}{1\cdot3}\right)\)\(\left(1+\frac{1}{2\cdot4}\right)\)\(\left(1+\frac{1}{3\cdot5}\right)\)\(......\left(1+\frac{1}{99\cdot101}\right)\)
\(=\frac{4}{1\cdot3}\)\(\cdot\frac{9}{2\cdot4}\)\(\cdot\frac{16}{3\cdot5}\)\(\cdot\cdot\cdot\cdot\cdot\frac{10000}{99\cdot101}\)
\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(=\frac{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99\cdot101}\cdot\frac{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}{3\cdot4\cdot5\cdot\cdot\cdot\cdot99}\)
\(=\frac{1}{101}\cdot200\)
\(=\frac{200}{101}\)
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Tính nhanh: \(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(B=\left(\frac{1}{1.3}+1\right).\left(\frac{1}{2.4}+1\right).\left(\frac{1}{3.5}+1\right).....\left(\frac{1}{99.101}+1\right)\)
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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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:
\(A=\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
\(B=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
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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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