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

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

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

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

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

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

\(=\dfrac{1}{39}+\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{\dfrac{117119}{1000}}+\dfrac{8}{39}\)

\(=\dfrac{1}{4641}-\dfrac{2852}{13923}-\dfrac{5000}{117119}+\dfrac{8}{39}\)

\(=-\dfrac{9827881}{232949691}\)

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:

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

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

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

Sửa đề: \(C=3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}+\dfrac{5}{119}-\dfrac{10}{117}\)

\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(1+1-\dfrac{1}{117}\right)\left(5+1-\dfrac{1}{110}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)

\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(2-\dfrac{1}{117}\right)\left(6-\dfrac{1}{119}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)

Đặt \(a=\dfrac{1}{117}\) và \(b=\dfrac{1}{119}\) ta có:

\(C=\left(3+a\right).\left(4+b\right)-\left(2-a\right)\left(6-b\right)+5b-10a\)

\(=12+3b+4a+ab-12+2b+6a-ab+5b-10a\)

\(=10b=10.\dfrac{1}{119}=\dfrac{10}{119}\)

10\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

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