đặt a=\(\dfrac{1}{135}\),b=\(\dfrac{1}{651}\)

ta có A=(2+a)b-\(\dfrac{3}{105}\)-\(\dfrac{650}{105.651}\)-4ab+\(\dfrac{12}{351}\)

A= (2+a) b-\(\dfrac{3}{105}\)-650.\(\dfrac{1}{105}\).\(\dfrac{1}{651}\)-4ab+12a

A=(2+a) b-\(\dfrac{9}{315}\)-650.\(\dfrac{3}{315}\).\(\dfrac{1}{651}\)-4ab+12a

A=(2+a)-b-9a-650.3a.b-4ab+12a

A=(2+a)b-9a-1950ab-4ab+12a=(2+a)b+3a-1954ab=-1953a+3a+2b

cách 2

Đặt \(\dfrac{1}{315}=a;\dfrac{1}{651}=b;\) thay vào \(A\) đc:

\(A=(2+a).b-\left(3+1-b\right).3a-4ab+12a\)

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

\(=2b=\dfrac{2}{651}\)

Đặt :

\(\dfrac{1}{315}=a;\dfrac{1}{651}=b\) thay vào A ta được :

\(A=\left(2+a\right)b-\left(3+1-b\right).3a-4ab+12a\)

\(\Leftrightarrow A=2b+ab-12a+3ab-4ab+12a\)

\(\Leftrightarrow A=2b\)

Thay \(b=\dfrac{1}{651}\) ta dc :

\(A=\dfrac{2}{651}\)

Đặt \(\dfrac{1}{315}=x,\dfrac{1}{651}=y\)

\(\Rightarrow A=\left(2+x\right)y-3x\left(4-y\right)-4xy+12x\)

\(=2y+xy-12x+3xy-4xy+12x\)

\(=2y\)

Thay \(y=\dfrac{1}{651}\Rightarrow A=\dfrac{2}{651}\)

Vậy...

Đặt S= \(2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315.651}+\dfrac{4}{105}\)

= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+\dfrac{651-1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)

= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+1-\dfrac{1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)

Đặt \(\dfrac{1}{315}=a,\dfrac{1}{651}=b\)

\(\Rightarrow S=\left(2+a\right).b-3a.\left(4-b\right)-4ab+12a\)

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

\(=2b=\dfrac{2}{651}\)