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