\(\frac{A}{4}=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\)

\(\frac{A}{4}=\frac{7-3}{3.7}+\frac{11-7}{7.11}+...+\frac{99-95}{95.99}\)

\(\frac{A}{4}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(A=\frac{4.32}{99}\)

\(4.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{95}-\frac{1}{99}\\ 4.A=\frac{1}{3}-\frac{1}{99}\\ 4.A=\frac{32}{99}\\ A=\frac{32}{99}:4\\ A=\frac{8}{99}\)

\(A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\)

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(A=\frac{1}{3}-\frac{1}{99}\)

\(A=\frac{32}{99}\)

Vậy \(A=\frac{32}{99}\)

A = 4/3 x 7 + 4/7 x 11 + 4/11 x 15 + .... + 4/95 x 99

A = 4/3 - 4/7 + 4/7 - 4/11 + 4/11 - 4/15 + ..... + 4/95 - 4/99

A = 4/3 - 4/99

A = 128/99

\(A=4\left(\frac{4}{3.7}+\frac{4}{7.11}+.......+\frac{4}{95.99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.......+\frac{1}{95}-\frac{1}{99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=4.\left(\frac{33}{99}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}=\frac{128}{99}\)

\(A=\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+...+\frac{4}{95}.\frac{4}{99}\)

\(A=4\left(\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+...+\frac{1}{95}.\frac{1}{99}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=4.\frac{32}{99}\)

\(A=\frac{128}{99}\)

Ta có:3/5+3/7-3/11=3.(1/5+1/7-1/11)

4/5+4/7-4/11=4.(1/5+1/7-1/11)

=>M=[3.(1/5+1/7-1/11)]/[4.(1/5+1/7-1/11)]=3/4

M = \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{3}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)

\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3 \left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)

=3[1/5+1/7-1/11]

-----------------------     =3/4

  4[1/5+1/7-1/11]

\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\cdot\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\cdot\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)

@_@

Học tốt ^^

\(M=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{13}{4}-\frac{13}{5}+\frac{13}{7}+\frac{13}{11}}\)

\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{13.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\)

\(=\frac{3}{13}\)

\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{91}\times\frac{4}{95}+\frac{4}{95}\times\frac{4}{99}\)

\(=4\left(\frac{1}{3\times7}+\frac{1}{7.11}+...+\frac{1}{91\times95}+\frac{1}{95\times99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{99}\right)=4\times\frac{32}{99}=\frac{128}{99}\)