\(y=\frac{1}{2.3}-\frac{2}{3.4}+\frac{3}{4.5}-...+\frac{99}{100.101}-\frac{100}{101.102}\)
Những câu hỏi liên quan
Tính tổng hoặc hiệu sau:Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+frac{1}{4.5}+..................+frac{1}{100.101}+frac{1}{101.102}Bfrac{1}{1.2}-frac{1}{2.3}-frac{1}{3.4}-frac{1}{4.5}- .....................-frac{1}{100.101}-frac{1}{101.102}
Đọc tiếp
Tính tổng hoặc hiệu sau:
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+..................+\(\frac{1}{100.101}\)+\(\frac{1}{101.102}\)
B=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\)-\(\frac{1}{4.5}\)- .....................-\(\frac{1}{100.101}\)-\(\frac{1}{101.102}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
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A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
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B=1/1.2-1/2.3-1/3.4-1/4.5-.......1/100.101-1/101.102
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/100-1/101+1/101-1/102
B=1-1/102
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\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^{^2}}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
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\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
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Tính tổng 1/2.3 - 2/3.4 + 3/4.5 +....+99/100.101 - 100/101.102
giúp mình
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{100^2}{100.101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4...100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3....100\right).\left(2.3.4...101\right)}\)
\(=\frac{1.1}{1.101}\)
\(=\frac{1}{101}\)
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\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}.....\frac{100^2}{100\cdot101}\)
\(=\frac{1.1}{1\cdot2}\cdot\frac{2.2}{2.3}\cdot\frac{3.3}{3.4}.....\frac{100.100}{100.101}\)
\(=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot100\right)\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot101\right)}\)
\(=\frac{1}{101}\)
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(de bai)=\(\frac{1^2.2^2.3^2...100^2}{1.2.2.3.3.4...100.101}\)
=\(\frac{1.1.2.2.3.3.4...100.100}{1.2.2.3.3.4...100.101}\)=\(\frac{1}{101}\)
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Xem thêm câu trả lời
a)\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)
\(=\frac{1}{1.101}\)
\(=\frac{1}{101}\)
k cho mk nha
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Tính:
a) \(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
b)\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{59^2}{58.60}\)
a,1^2/1.2 . 2^2/2.3 . 3^2/3.4 ... 99^2/99.100 . 100^2/100.101
= 1/2 . 2/3 . 3/4 ... 99/100 . 100/101
=( 2.3.4....100/2.3.4...100) . 1/101
= 1 . 1/101
=1/101
ý b tương tự nhé !
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Bạn kia đéo biết thì thôi xen vào làm gì?
Xem thêm câu trả lời
Tính
\(\frac{1^2}{1.2}+\frac{2^2}{2.3}+\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
Thực hiện phép tính :
a)A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{2017}}{2017+\frac{2016}{2}+\frac{2015}{3}+.............\frac{1}{2016}}\)
b)B=\(\frac{-1^2}{1.2}.\frac{-2^2}{2.3}+..........+\frac{-100^2}{100.101}.\frac{-101^2}{101.102}\)
Giúp mình vs
Tính giá trị biểu thức:
M=\(\left(1-\frac{2}{2.3}\right)\).\(\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{101.102}\right)\)
GIÚP MIK NHA MIK ĐG CẦN GẤP.CẢM ƠN!!
Xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{2}{2.3}=\frac{1.4}{2.3}\) ; \(1-\frac{2}{3.4}=\frac{2.5}{3.4}\) ; ... ; \(1-\frac{2}{101.102}=\frac{100.103}{101.102}\)
\(\Rightarrow M=\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\cdot\cdot\frac{100.103}{101.102}\)
\(M=\frac{\left(1.2...100\right).\left(4.5...103\right)}{\left(2.3...101\right).\left(3.4...102\right)}=\frac{103}{101.3}=\frac{103}{303}\)
Vậy \(M=\frac{103}{303}\)