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

a: \(\dfrac{119}{117}=1+\dfrac{2}{117}\)

\(\dfrac{117}{115}=1+\dfrac{2}{115}\)

mà 2/117<2/115

nên \(\dfrac{119}{117}< \dfrac{117}{115}\)

hay \(-\dfrac{119}{117}>-\dfrac{117}{115}\)

b: \(\dfrac{-22}{35}=\dfrac{-22\cdot177}{35\cdot177}=-\dfrac{3894}{6195}\)

\(\dfrac{-103}{177}=\dfrac{-103\cdot35}{177\cdot35}=\dfrac{-3605}{6195}\)

mà -3894<-3605

nên -22/35<-103/177

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

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

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

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

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

\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)

\(\Leftrightarrow\dfrac{x+110}{113}+\dfrac{x+110}{115}-\dfrac{x+110}{117}-\dfrac{x+110}{119}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)

Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)

\(\Rightarrow x+110=0\)

\(\Rightarrow x=-110\)

\(\dfrac{x-3}{133}+\dfrac{x-5}{155}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)

\(\Leftrightarrow\left(\dfrac{x-3}{113}+1\right)+\left(\dfrac{x-5}{115}+1\right)=\left(\dfrac{x-7}{117}+1\right)+\left(\dfrac{x-9}{119}+1\right)\)

\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}=\dfrac{x+130}{117}+\dfrac{x+130}{119}\)

\(\Leftrightarrow\dfrac{x+130}{113}+\dfrac{x+130}{115}-\dfrac{x+130}{117}-\dfrac{x+130}{119}=0\)

\(\Leftrightarrow\left(x+130\right)\left(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\right)=0\)

Mà \(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\ne0\)

\(\Leftrightarrow x+130=0\)

\(\Leftrightarrow x=-130\)

Vậy..

\(tuA=1003+1007+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{2010}{119}=2010\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\)\(mauA=1003+1008+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{2011}{119}=2011\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\)có \(\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\ne0=>A=\dfrac{2010}{2011}\)

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 : \(\dfrac{1}{117}\) = 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 = \(\dfrac{10}{117}\)