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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
\(B=\left(2+\dfrac{2}{315}\right).\dfrac{1}{651}-\dfrac{1}{105}.\left(3+1-\dfrac{1}{651}\right)-4.\dfrac{1}{315}.\dfrac{1}{651}+\dfrac{4}{105}\)\(=\left(2+\dfrac{2}{3.105}\right).\dfrac{1}{651}-\dfrac{1}{105}.\left(4-\dfrac{1}{651}\right)-4.\dfrac{1}{3.105}.\dfrac{1}{651}+\dfrac{4}{105}\)\(=\left(2+\dfrac{2}{3}.\dfrac{1}{105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(4-\dfrac{1}{651}\right)-\dfrac{4}{3}.\dfrac{1}{105}.\dfrac{1}{651}+4.\dfrac{1}{105}\)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
Tham khảo ở link này nè:
Câu hỏi của Chuotconbebong2004 - Toán lớp 8 | Học trực tuyến
Chúc bn học giỏi!
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 :
\(A=\dfrac{2}{651}\)
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 :>