Question

Tính giá trị biểu thức

\(A=3\dfrac{1}{117}.\dfrac{1}{119}-\dfrac{4}{117}.5\dfrac{118}{119}-\dfrac{5}{117.119}+\dfrac{8}{39}\)

help me ~~

Answer 1

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

Answer 2

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

Answer 3

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