\(\frac{534.534-96}{534.533+438}=\frac{534.\left(533+1\right)-96}{534.533+438}=\frac{534.533+534-96}{534.533+438}\frac{534.533+438}{534.533+438}\)\(=1\)

Chính xác là bằng 1

Bn ơi cho mình lời giải cụ thể với

=438.(1/111.113+1/113.113+...+1/217.219)

=438.1/2.(2/111.113+2/113.115+...+1/217.219)

=438.1/2.(1/111-1/113+1/113-1/115+...+1/217-1/219)

=438.1/2.(1/111+0+0+..+0-1/219)

=438.1/2.(1/111-1/219)=219/111-1=

b tự tính

 \(A=\frac{438}{111.113}+\frac{438}{113.115}+\frac{438}{115.117}+...+\frac{438}{217.219}\)

\(A=219\left(\frac{2}{111.113}+\frac{2}{113.115}+\frac{2}{115.117}+...+\frac{2}{217.219}\right)\)

\(A=219\left(\frac{1}{111}-\frac{1}{113}+\frac{1}{113}-\frac{1}{115}+\frac{1}{115}-\frac{1}{117}+...+\frac{1}{217}-\frac{1}{219}\right)\)

\(A=219\left(\frac{1}{111}-\frac{1}{219}\right)\)

642 + 455 + 438 + 675= (642+438)+(455+675)

                                    = 1080+ 1130

                                     = 2210

Hôctot

642 + 455 + 438 + 675

=(455+675)+(642+438)

=1130+1080

=2210

\(642+455+438+675\)

\(=1080+1130\)

\(=2210\)

Đặt bt bằng A

Ta có 2A= 2(2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96 + 2/192)

2A= 4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96

A= 2A-A= (4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96) - (2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96  2/192)

A=4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96 - 2/3 - 2/6 - 2/12 - 2/24 - 2/48 - 2/96 - 2/192

A=(2/3 - 2/3) + (2/6 - 2/6) + ( 2/12 -  2/12) + (2/24 - 2/24) + (2/48 - 2/48) + ( 2/96 -  2/96) + (4/3 - 2/192)

A=0+0+0+0+0+0+ (256/192 - 2/192)

A=254/192

A=127/96(rút gọn phân số) 

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{192}\)

\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{192}\right)\)

\(=2\times\frac{191}{192}\)

\(=\frac{382}{192}=\frac{191}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2.\left(1-\frac{1}{192}\right)\)

\(=2.\frac{191}{192}\)

\(=\frac{382}{192}=\frac{191}{96}\)

Câu 1

Ta có \(\frac{119x83-183}{120x83-266}=\frac{119x83-183}{119x83+83-266}=\frac{119x83-183}{119x83-183}=1\) 

\(A=\frac{3-2}{2\times3}+\frac{5-3}{3\times5}+\frac{8-5}{5\times8}+...\frac{38-30}{30\times38}+\frac{47-38}{38\times47}\)

\(A=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{5}{3\times5}-\frac{3}{3\times5}+...\frac{38}{30\times38}-\frac{30}{30\times38}+\frac{47}{38\times47}-\frac{38}{38\times47}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{30}-\frac{1}{38}+\frac{1}{38}-\frac{1}{47}\)

\(A=\frac{1}{2}-\frac{1}{47}=\frac{47}{94}-\frac{2}{94}=\frac{45}{94}\)