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)

\(=\frac59\cdot P=\frac59\cdot\frac{10\left(10^{n}-1\right)-9n}{9}=\frac{5\left\lbrack10\left(10^{n}-1\right)-9n\right\rbrack}{81}\)

=\(\frac{5}{9}\)

\(\frac{5+55+555+5555}{9+99+999+9999}=\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\frac{5}{9}\)

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 !!!!!!!!!!!!!

\(3A=9+99+....+99....99=10-1+10^2-1+...+10^{50}-1\)

\(=\left(10+10^2+...+10^{50}\right)-50\)

Đến đây dễ hơn rồi.

\(\frac{9}{5}B=9+99+...+99.99\)tương tự A

C tương tự A.

tự làm

hoi vo van

đúng òi vớ va vớ vẩn

746.3125

6567,27 

\(\frac{55^{12}.9^{11}}{99^{12}.5^{11}}\)

\(=\frac{\left(5.11\right)^{12}.9^{11}}{\left(9.11\right)^{12}.5^{11}}\)

\(=\frac{5^{12}.11^{12}.9^{11}}{9^{12}.11^{12}.5^{11}}\)

\(=\frac{5}{9}\)