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Câu hỏi của Chuotconbebong2004 - Toán lớp 8 | Học trực tuyến
Chúc bn học giỏi!
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
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 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 \(\dfrac{1}{315}=a,\dfrac{1}{651}=b\)
\(\Rightarrow S=\left(2+a\right).b-3a.\left(4-b\right)-4ab+12a\)
\(=2b+ab-12a+3ab-4ab+12a\)
\(=2b=\dfrac{2}{651}\)
a: \(=\dfrac{\left(2\cdot547+1\right)\cdot3}{547\cdot211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{2735}{547\cdot211}=\dfrac{5}{211}\)
b: x=7 nên x+1=8
\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)
\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}-...-x^3-x^2+x^2+x-5\)
=x-5=7-5=2