61 phan 96

\(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+\(\frac{1}{24}\)+\(\frac{1}{48}\)+\(\frac{1}{96}\) 

= ( \(\frac{1}{3}\)+\(\frac{1}{6}\)) + ( \(\frac{1}{12}\)+ \(\frac{1}{24}\)) + ( \(\frac{1}{48}\)+\(\frac{1}{96}\))

=       \(\frac{1}{2}\)     +         \(\frac{1}{8}\)         +          \(\frac{1}{32}\) 

=                 \(\frac{5}{8}\)        +          \(\frac{1}{32}\) 

=                             \(\frac{21}{32}\)

cái a bằng 1962

cái b bằng 127/192

à quên mình chưa rút gọn phân số đấy đâu bạn ạ

ban rút gọn phân số đấy hộ mình nha

bạn giải từng bước ra giúp mình nhé

\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)

\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

\(B=\frac{3}{4}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)

\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)

\(A=\frac{2}{3}-\frac{1}{192}\)

\(A=\frac{127}{192}\)

\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

      \(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)

      \(C=\frac{1990.997}{1994.995}\)

      \(C=\frac{995.2+997}{997.2+995}=1\)

\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)

\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)

Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{96}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+.....+\frac{1}{192}\)

\(\Rightarrow A-\frac{1}{2}A=\frac{1}{3}-\frac{1}{192}\)

\(\Rightarrow\frac{1}{2}A=\frac{21}{64}\)

\(\Rightarrow A=\frac{21}{64}.2=\frac{21}{32}\)

\(\frac{95}{96}\)ko lý do

Đặt biểu thức trên là A ta có:

A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)

A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)

A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)

A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)

A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)

bằng 1

Câu 1

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