\(\frac{16}{11},-\frac{5}{9},\frac{10}{539}\)

\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3}{1}=8\)

Vậy A = 8

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\frac{8}{33}=\frac{10}{11}.\frac{33}{8}\)

\(=\frac{15}{4}\)

Trả lời:

\(\frac{10}{11}\div\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}\div\frac{8}{33}\)

\(=\frac{10}{11}\times\frac{33}{8}\)

\(=\frac{15}{4}\)

\(\frac{10}{11}:(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11})\)

=\(\frac{10}{11}:(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11})\)

=\(\frac{10}{11}:(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\)

=\(\frac{10}{11}:(\frac{1}{3}-\frac{1}{11})\)

=\(\frac{10}{11}:\frac{8}{33}\)

= \(\frac{10}{11}.\frac{33}{8}\)= \(\frac{15}{4}\)

\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=\frac{10}{11}:\frac{8}{33}\)

\(=\frac{15}{4}\)

Học tốt

\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\frac{8}{33}\)

\(=\frac{10}{11}.\frac{33}{8}\)

\(=\frac{15}{4}\)

\(\frac{10}{11}:\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\frac{8}{33}=\frac{15}{4}\)

\(=\frac{2^3.3^3.5^3.2^3.7}{3.2^4.5^3.2.7}=\frac{2^6.5^3.3^3}{3.2^5.5^3}=2.3^2=2.9=18\)

=\(\frac{2^3\cdot3^3\cdot5^3\cdot2^3\cdot7}{3\cdot2^4\cdot5^3\cdot2\cdot7}=\frac{2^{3_{ }}\cdot3^3\cdot2^3}{3\cdot2^4\cdot2}=\frac{1728}{96}=\frac{18}{1}=18\)

2 câu trên câu nào cũng đúng. nhưng làm theo cách 2 thì nhanh hơn

= (-2)^3. 3= -24

\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.5^3.2^4.42}\)

\(=\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3.5^3.2^4.2.3.7}=\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3^2.5^3.2^5.7}=\frac{-2.3}{1}=-6\)

học tốt~~~

\(\frac{\left(-2^3\right).3}{2^2}=\frac{-24}{4}=-6\)

Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)

\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2.2.2.5.7.5.5.7.7.7}{2.5.7.7.2.5.7.7}=\frac{2.5}{1}=10\)

Ko biết có đúng ko

\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2^3.\left(5.5^2\right).\left(7.7^3\right)}{2^2.5^2.7^{2^2}}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)