\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)

\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)

\(=\frac{5}{8}:\frac{15}{16}\)

\(=\frac{2}{3}\)

2/3

Tk mk nha

a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)

=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có nghiệm là x = 2004 .

b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)

=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)

=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)

=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình có nghiệm là x = 100 .

Bài này bạn tính thôi

Trình bày ra 

Rồi làm thôi

Áp dụng công thức tính phân số ở Tiểu học

\(\frac{5\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}\div\frac{15\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}=\frac{5}{8}\div\frac{15}{16}=\frac{2}{3}\)

F=5-5x(1/3+1/9-1/27) /8-8x(1/3+1/9-1/27)

: 15-15x(1/11+1/121) /16-16x(1/11+1/121)

=5-5x1/8-8x1

: 15-15x1/16-16x1

=0:0=0

chắc vậy!

\(A=\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}\)

\(\Rightarrow A=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}\)

\(\Rightarrow A=\dfrac{5}{8}\)

\(=\frac{5-\frac{5}{3}-\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}-\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)

\(=\frac{5\left(1-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)

\(=\frac{5}{8}:\frac{15}{16}=\frac{5}{8}.\frac{16}{15}=\frac{2}{3}\)