\(C=\frac{631}{315}.\frac{1}{651}-\frac{1}{103}.\frac{2603}{651}-\frac{4}{315}.\frac{1}{651}+\frac{4}{105}\)

    \(=\frac{1}{651}.\left(\frac{631}{315}-\frac{4}{315}\right)+\frac{2603}{68355}+\frac{4}{105}\)

    \(=\frac{1}{651}.\frac{209}{105}+\frac{2603}{68355}+\frac{4}{105}\)

     \(=\frac{1}{105}.\left(\frac{209}{651}+\frac{4}{105}\right)+\frac{2603}{68355}\)

     \(=105.\frac{167}{465}+\frac{2603}{68355}\)

     \(=\frac{1169}{31}+\frac{2603}{68355}=37,74775803\)

Tính nhanh và gọn hết cỡ đc có vậy thôi. Bạn xem lại đề bài nhé 

bài này có thể đặt số= 1 ẩn

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

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

Đặt \(\frac{1}{315}=x;\frac{1}{651}=y\),khi đó C trở thành:

\(C=\left(2+x\right).y-3x.\left(4-y\right)-4xy+12y\)

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

\(C=2y=2.\frac{1}{651}=\frac{2}{651}\)

Vậy C=2/651

\(\frac{2}{651}\)

Đặt \(a=\frac{1}{315}\),    \(b=\frac{1}{651}\)ta có :

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

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

\(\Rightarrow A=2b=\frac{2}{651}\)

Đặt H = \(2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}-3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)ư

=> H = \(\frac{1}{315}\cdot\frac{1}{651}+2\frac{1}{651}-\frac{1}{105}\cdot\left(4-\frac{1}{651}\right)-\frac{4}{135}\cdot\frac{1}{651}+\frac{4}{105}\)

=> H = \(\frac{1}{135}\cdot\frac{1}{651}+2\frac{1}{651}-\frac{4}{105}+\frac{1}{105}\cdot\frac{1}{651}-\frac{4}{135}\cdot\frac{1}{651}+\frac{4}{105}\)

=> H = \(\frac{1}{651}\left(\frac{1}{315}+\frac{1}{105}+2-\frac{4}{315}\right)+\frac{4}{105}-\frac{4}{105}\)

=> H = \(\frac{2}{651}\)

Vậy H = \(\frac{2}{651}\)