ghê nha, hỏi bài của thầy phúc lun!

các bn không cần giải nữa nhé,mik tìm đc cách giải r.....còn ai giải r thì mik cảm ơn và sẽ k cho mỗi ngày nhé.

8489/120

\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

chỗ giữa 2 phân số là dấu nhân hay sao mà chả thấy dấu j thế?

Trà My:là dấu nhân thế cx hỏi

a,Gọi tổng trên là A.

Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)

b,Gọi tổng trên là B

Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)

\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)

Đặt A=B*C

B=\(\frac{24\cdot47-23}{24+47-23}=\frac{1128-23}{71-23}=\frac{1105}{48}\)

C=\(\frac{3\cdot\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\cdot\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{1}{3}\)

Suy ra A =\(\frac{1105}{144}\)

24.47 - \(\frac{23}{24}\)+ 47 - 23

= 24.47 - 0.9583333333 + 47 - 23

= 23.51166667 + 47 - 23

= 70.51166667 - 23

= 47.51166667

E xloi !; E giải đc mỗi bài 2 thui ạ !

giải bài 2 cx đc ko sao

\(\frac{24.27-23}{24+47.23}=\frac{648-23}{24+1081}=\frac{625}{1105}=\frac{125}{221}\)

_Học_tốt_

=^.^=

\(\frac{24.27-23}{24+47.23}\\ =\frac{\left(23+1\right).27-23}{23+1+47.23}\\ =\frac{23.27+27-23}{23.\left(47+1\right)+1}\\ \)

\(\frac{24.27-23}{24+47.23}=\frac{648-23}{24+1081}=\frac{625}{1105}=\frac{125}{221}\)

Bài 1: 

\(A=\frac{24.47-22}{24+47.23}.\frac{5+\frac{5}{7}+\frac{5}{11}-\frac{5}{13}+\frac{5}{1001}}{6+\frac{6}{7}+\frac{6}{11}-\frac{6}{13}+\frac{6}{1001}}\)\(=\frac{47.23+47-22}{24.47.23}.\frac{5\left(1+\frac{1}{7}+\frac{1}{11.}-\frac{1}{13}+\frac{1}{1001}\right)}{6\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}+\frac{1}{1001}\right)}\) 

                                                                         \(=\frac{47.23+24}{24+47.23}.\frac{5}{6}\) 

                                                                         \(=1.\frac{5}{6}=\frac{5}{6}\) 

Bài 2: 

\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\) 

                                      \(=3^{28}-3^{27}-3^{26}\) 

                                      \(=3^{22}\left(3^6-3^5-3^4\right)\) 

                                      \(=3^{22}.405\) chia hết cho 405 

 =>đpcm