\(A=3.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)

\(A=3.\left(1-\frac{1}{105}\right)\)

\(A=3.\frac{104}{105}\)

\(A=\frac{104}{35}\)

Em yêu cầu bác nhìn xuống dưới và bác sẽ biết cách làm 

Bác thấy rồi mà còn đăng

Thay số mà làm nhé

:))

Để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của các mẫu lên tử 

\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)

\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)

\(A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)

\(A=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{1}{93.97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\left(1-\frac{1}{97}\right)\)

\(A=\frac{1}{4}.\frac{96}{97}=\frac{24}{97}\)

cau nay chinh con me no

con nay la 1209876

111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000%

(1+2+3-4-5-6)+(7+8+9-10-11-12)+....+(97+98+99-100-101-102)=-27+-27+...+-27

                                                                                          =-27*17=-153

neu dug thi tk nha

*cam tớ hơi mờ nên thông cảm nhé:<.

\(\dfrac{12}{1\cdot5}+\dfrac{12}{5\cdot9}+...+\dfrac{12}{97\cdot101}\)

\(=3\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{97\cdot101}\right)\)

\(=3\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

=3*100/101

=300/101

           \(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

\(A=\frac{12}{1.5}+\frac{12}{5.9}+\frac{12}{9.13}+.............+\frac{12}{101.105}\)

     \(=3.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+............+\frac{4}{101.105}\right)\)

       \(=3\left(1-\frac{1}{105}\right)\)

          \(=3.\frac{104}{105}=\frac{312}{105}\)

a: \(=\dfrac{2^4\cdot3^6\cdot2\cdot3}{2^4\cdot3^6}=6\)

b: \(=\dfrac{2^{20}\cdot3^{20}}{2^{18}\cdot3^{18}}=2^2\cdot3^2=36\)

c: \(=\dfrac{12^5\cdot13}{12^6\cdot13}-\dfrac{12^8\cdot\left(-11\right)}{12^9\cdot\left(-11\right)}=\dfrac{1}{12}-\dfrac{1}{12}=0\)

-97-15+44-35-12+98 = - 17