Đặt a=315; b=651

\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)

\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)

\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)

\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)

=2/651

Đặt a=315; b=651

\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)

\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)

\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)

\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)

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

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

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

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

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 :

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

Đề đúng nhé bạn :

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

\(\Leftrightarrow B=\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}\left(3+\dfrac{650}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)

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

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

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

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

\(\Leftrightarrow B=2b+\left(ab+3ab-4ab\right)+\left(12a-12a\right)\)

\(\Leftrightarrow B=2b\)

\(\Leftrightarrow B=2.\dfrac{1}{651}\)

\(\Leftrightarrow B=\dfrac{2}{651}\)

Vậy \(B=\dfrac{2}{651}\)

:D