so sánh 2 psố sau:
\(\frac{6363}{8181}\) và \(\frac{5+55+555}{9+99+999}\)
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\(A=\frac{5+55+555+5555}{9+99+999+9999}=\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\frac{5}{9}\)
Tớ làm luôn
A=\(\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}\)
A=\(\frac{5}{9}\)
\(\frac{5+55+555+5555}{9+99+999+9999}=\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\frac{5}{9}\)\(\frac{5}{9}\)
dấu chấm bạn cứ hiểu là nhân nhé !
\(\frac{5+55+555+5555}{9+99+999+9999}=\frac{5+11x5+111x5+1111x5}{9+11x9+111x9+1111x9}\)\(=\frac{5x5x5x5}{9x9x9x9}=\frac{625}{6561}\)
Ta có: \(P=9+99+999+\cdots+99\ldots9\) (n chữ số 9)
\(=10-1+10^2-1+\cdots+10^{n}-1\)
\(=\left(10+10^2+\cdots+10^{n}\right)-n\)
Đặt \(A=10+10^2+\cdots+10^{n}\)
=>\(10A=10^2+10^3+\cdots+10^{n+1}\)
=>\(10A-A=10^2+10^3+\cdots+10^{n+1}-10-10^2-\cdots-10^{n}\)
=>\(9\cdot A=10^{n+1}-10=10\left(10^{n}-1\right)\)
=>\(A=\frac{10\left(10^{n}-1\right)}{9}\)
Ta có: \(P=\left(10+10^2+\cdots+10^{n}\right)-n\)
\(=\frac{10\left(10^{n}-1\right)}{9}-n=\frac{10\left(10^{n}-1\right)-9n}{9}\)
b: \(Q=5+55+555+\cdots+55\ldots5\) (n chữ số 5)
\(=\frac59\left(9+99+\cdots+999\ldots9\right)\) (n chữ số 9)
\(=\frac59\cdot P=\frac59\cdot\frac{10\left(10^{n}-1\right)-9n}{9}=\frac{5\left\lbrack10\left(10^{n}-1\right)-9n\right\rbrack}{81}\)
\(A=\frac{8\cdot4\cdot125\cdot25+96524+2476}{10\cdot125\cdot4\cdot25\cdot8}\)
\(A=\frac{1+96524+2476}{10}\)
\(A=\frac{99001}{10}\)
\(B=\frac{5+55+555+5555}{9+99+999+9999}\)
\(B=\frac{9}{5}\)
\(\frac{5+55+555+5555}{9+99+999+9999}=\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\frac{5}{9}\)
a ) 9 + 99 + 999 + ........ + 999...99 (có 10 chữ số 9)
= (10 - 1) + (100 - 1) + (1000 - 1) + ..... + (100...000 - 1)
= (10 + 100 + 1000 + .... + 100...00 ) - (1 + 1 + .... + 1)
= 1111....1110 - 10
= 1111....1100
Các ý khác tương tự nha !!!!!!!!!!!!!
\(\frac{5+55+555+5555}{9+99+999+9999}=\)
\(=\frac{5x\left(1+11+111+1111\right)}{9x\left(1+11+111+1111\right)}=\frac{5}{9}\)
ê thằng kia làm kiểu j