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

\(B=\left(2+\dfrac{2}{315}\right).\dfrac{1}{651}-\dfrac{1}{105}.\left(3+1-\dfrac{1}{651}\right)-4.\dfrac{1}{315}.\dfrac{1}{651}+\dfrac{4}{105}\)\(=\left(2+\dfrac{2}{3.105}\right).\dfrac{1}{651}-\dfrac{1}{105}.\left(4-\dfrac{1}{651}\right)-4.\dfrac{1}{3.105}.\dfrac{1}{651}+\dfrac{4}{105}\)\(=\left(2+\dfrac{2}{3}.\dfrac{1}{105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(4-\dfrac{1}{651}\right)-\dfrac{4}{3}.\dfrac{1}{105}.\dfrac{1}{651}+4.\dfrac{1}{105}\)Đặt \(a=\dfrac{1}{105};b=\dfrac{1}{651}\)

\(B=\left(2+\dfrac{2}{3}a\right)b-a\left(4-b\right)-\dfrac{4}{3}ab+4a\)

\(=2b+\dfrac{2}{3}ab-4a+ab-\dfrac{4}{3}ab+4a\)

\(=2b+\dfrac{1}{3}ab\)

Còn lại bạn tự làm nha, mình ko chắc lắm, nếu sai thông cảm nha

1/ \(4x\left(x-1\right)-x\left(4x-2\right)=5\left(x-3\right)\)

\(\Leftrightarrow4x^2-4x-4x^2+2x=5x-15\)

\(\Leftrightarrow-2x=5x-15\)

\(\Leftrightarrow7x=-15\)

\(\Leftrightarrow x=-\dfrac{15}{7}\)

Vậy ..

2/ Đặt : \(\dfrac{1}{105}=x;\dfrac{1}{651}=y\left(x,y>0\right)\)

Ta có :

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

\(=\left(2+\dfrac{1}{3.105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(3+\dfrac{651-1}{651}\right)+\dfrac{4}{105}\)

\(=\left(2+\dfrac{1}{3}.\dfrac{1}{105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(4-\dfrac{1}{651}\right)-\left(\dfrac{3}{3.105.651}+\dfrac{1}{3.105.651}\right)+\dfrac{4}{105}\)

\(=\left(2+\dfrac{1}{3}x\right)y-x\left(4-y\right)-xy-\dfrac{1}{3}xy+4x\)

\(=2y+\dfrac{1}{3}xy-4x+xy-xy-\dfrac{1}{3}xy+4x\)

\(=2y\)

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

Có sửa lại đề 1 chút á :>

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

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

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

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

Tham khảo ở link này nè:

Câu hỏi của Chuotconbebong2004 - Toán lớp 8 | Học trực tuyến

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