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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 !!!!!!!!!!!!!
\(A=\frac{5+55+555+5555}{9+99+999+9999}=\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\frac{5}{9}\)
Tớ làm luôn
A=\(\frac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}\)
A=\(\frac{5}{9}\)
a, \(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
b, \(\dfrac{5+55+555+5555}{9+99+999+9999}\)
\(=\dfrac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\dfrac{5}{9}\)
Chúc bạn học tốt!!!
a,\(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
\(\frac{5+55+555+5555}{9+99+999+9999}=\frac{5\cdot\left(1+11+111+1111\right)}{9\cdot\left(1+11+111+1111\right)}=\frac{5}{9}\)
Ta có:
\(C= 4+44+444+......+4444444444\)
\(C= 4.(10.1+9.10+8.100+7.1000+...+1.1000000000\)
\(C= 4.(100+90+800+7000+60000+500000+4000000+30000000+200000000+1000000000)\)
\(C=4.12345678900\)
\(C=4938271600\)
Tương tự.
A=(10-1)+(100-1)+(1000-1)+....+(100...000-1)
50 chữ số 0=(10+100+1000+100...000)-(1+1+1+..+1+1)
50 chữ số 0 ; 50 chữ số 1
=111...1110-50
50 chữ số 1
=111...111060
9 chữ số 1
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)
\(=\frac59\cdot P=\frac59\cdot\frac{10\left(10^{n}-1\right)-9n}{9}=\frac{5\left\lbrack10\left(10^{n}-1\right)-9n\right\rbrack}{81}\)