Sễ thui:

 A = 1 - 1/3 + 1/3 - 1/5 + ........ + 1/2011 - 1/2013

    = 1- 1/2013

    = 2012/2013

B = 5/2 + 3/10 + 

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)

\(A=1-\frac{1}{2013}\)

\(A=\frac{2012}{2013}\)

phần B nhân hạng tử 1 số nào đó tự tìm nhe ^^

Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được

tôi biết câu này nè

LÀM TẮT NHÉ :

\(C=\frac{3}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\right)\))

\(D=\frac{4}{5}\left(\frac{1}{2}-\frac{1}{7}+...+\frac{1}{102}-\frac{1}{107}\right)\)

tương tự với các phần còn lại

b) \(\frac{-5\cdot7^5+7^4}{7^6\cdot10-2\cdot7^5}\)

\(=\frac{-35\cdot7^4+7^4}{7^5\cdot70-2\cdot7^5}\)

\(=\frac{7^4\left(-35+1\right)}{7^5\left(70-2\right)}\)

\(=\frac{7^4\cdot\left(-34\right)}{7^5\cdot68}\)

\(=\frac{-1}{14}\)

Chắc sai =))

S = \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+.......+\frac{5}{17.19}\)

S : 5 = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{17.19}\) 

S : 5 = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}+.......+\frac{1}{17}-\frac{1}{19}\)

=> S : 5 = 1 - \(\frac{1}{19}=\frac{19}{19}-\frac{1}{19}=\frac{18}{19}\)

=> S = \(\frac{18}{19}x5=\frac{90}{19}\)

=5/19

mình cũng hỏi bài ni này

1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25

=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25

=> Kết quả là: 1/5 - 1/25 = 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101

=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101

=> kết quả là 2/1 - 2/101 =200/101

a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

=\(\frac{1}{5}-\frac{1}{25}\)

=\(\frac{4}{25}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)

=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)

=\(2.\frac{100}{101}\)

=\(\frac{200}{101}\)

a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{5}{25}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

a,2.7^x+1-3.7^x=77

=>2.7^x.7¹-3.7^x=77

=>7^X.(2.7-3)=77

=>7^x.11=77

=>7^x=7

=>7^x=7¹

=>x=1

TA CÓ : \(\frac{2^3.3^4}{2^2.3^2.5}\)= \(\frac{2^3.3^4}{\left(2.3\right)^2.5}\)= \(\frac{2^3.3^4}{6^2.5}\)= \(\frac{2^3.3^4}{36.5}\)= \(\frac{8.81}{180}\)=  \(\frac{648}{180}\)= 648 : 180 = 3,6 HOẶC \(\frac{648}{180}\)= \(\frac{18}{5}\)

câu b) thì sao bạn

ngu..............................

tao lao

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)

\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)

\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)