Ta có: \(A=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{352}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{352-2852-5}{117\cdot119}+\dfrac{8}{39}\)

\(=\dfrac{-835}{4641}+\dfrac{8}{39}\)

\(=\dfrac{3}{119}\)

Đặt \(\dfrac{1}{117}=x;\dfrac{1}{119}=y\)

\(\Rightarrow\dfrac{1}{39}=3x\)

Ta có: \(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+8.3x\)

\(=3y+xy-20x-4x+4xy-5xy+24x\)

\(=3y\)

Thay \(y=\dfrac{1}{119}\rightarrow A:\)

\(A=3.\dfrac{1}{119}=\dfrac{3}{119}\)

Vậy \(A=\dfrac{3}{119}.\)

Đặt \(a=\dfrac{1}{117};b=\dfrac{1}{119}\) thay vào A được:

A=\(\left(3+a\right)b-4a\left(6-b\right)-5ab+\dfrac{8}{39}\)

=\(3b+ab-24a+4ab-5ab+\dfrac{8}{39}\)

=\(3b-24a+\dfrac{8}{39}\) (1)

Thay \(a=\dfrac{1}{117};b=\dfrac{1}{119}\) vào (1) ta đuợc:

A=\(\dfrac{3}{119}-\dfrac{24}{117}+\dfrac{8}{39}=\dfrac{3}{119}-0=\dfrac{3}{119}\)

Chúc các bn học tốt

\(\dfrac{3}{119}\)

Đặt 117=a; 119=b

Theo đề, ta có:

\(B=\left(3+\dfrac{1}{a}\right)\cdot\dfrac{1}{b}-\dfrac{4}{a}\cdot\left(5+\dfrac{b-1}{b}\right)-\dfrac{5}{a\cdot b}+8:\dfrac{a}{3}\)

\(=\dfrac{3a+1}{a}\cdot\dfrac{1}{b}-\dfrac{4}{a}\cdot\dfrac{5b+b-1}{b}-\dfrac{5}{ab}+\dfrac{24}{a}\)

\(=\dfrac{3a+1-24b+4-5}{ab}+\dfrac{24}{a}=\dfrac{3a-24b+24b}{ab}=\dfrac{3a}{ab}=\dfrac{3}{b}=\dfrac{3}{119}\)

Đặt \(a=\dfrac{1}{117};b=\dfrac{1}{119}\)

Ta có:\(3\dfrac{1}{117}.\dfrac{1}{119}-\dfrac{4}{117}.5\dfrac{118}{119}-\dfrac{5}{117.119}+\dfrac{8}{39}\)

=\(\left(3+a\right)b-4a\left(6-b\right)-5ab+24a\)

\(\text{ }\)=3b+ab-24a+4ab-5ab+24a

\(=3b=3.\dfrac{1}{119}=\dfrac{3}{119}\)

Đặt : \(\dfrac{1}{117}\) = x ; \(\dfrac{1}{119}\) = y .

A = ( 3 + x)( 4 + y) - (1 + 1 - x)(5 + 1 - y) - 5y

<=> A = 12 + 3y + 4x + xy - ( 2 - x)( 6 - y) - 5y

<=> A = 12 + 3y + 4x + xy - 12 + 2y + 6x - xy - 5y

<=> A = 10x

<=> A = \(\dfrac{10}{117}\).

Vậy A = \(\dfrac{10}{117}\)