A = 11¹ + 11² + 11³ + ... + 11⁹⁹ + 11¹⁰⁰

= ( 11¹ + 11² ) + ( 11³ + 11⁴ ) + ( 11⁵ + 11⁶ ) + ..... + ( 11⁹⁹ + 11¹⁰⁰ )

= 11 ( 1 + 11 ) + 11³ ( 1 + 11 ) + 11⁵ ( 1 + 11 ) + .... + 11⁹⁹ ( 1 + 11 )

= ( 1 + 11 ) ( 11 + 11³ + 11⁵ + .... + 11⁹⁹ )

= 12 ( 11 + 11³ + 11⁵ + .... + 11⁹⁹ ) chia hết cho 12

Ta có \(11^1+11^2+11^3+...+11^{99}+11^{100}=\left(11^1+11^2\right)+\left(11^3+11^4\right)+..+\left(11^{99}+11^{100}\right)\)

\(=\left(11^1+11^2\right)+11^2.\left(11^1+11^2\right)+..+11^{98}.\left(11+11^2\right)\)

\(=132+11^2.132+...+11^{98}.132\)

\(=132.\left(11^0+11^2+...+11^{98}\right)\)

Có \(132⋮12\)nên \(132.\left(11^0+11^2+...+11^{98}\right)⋮12\)

Vậy \(11^1+11^2+11^3+...+11^{99}+11^{100}⋮12\)

\(=\left(11^1+11^2\right)+...+\left(11^{99}+11^{100}\right)\)

=11(1+11)+....+11^99(1+11)

=12(11+11^3+...+11^99)\(⋮\)12

\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)

có : `1-1/(11^99)<1`

\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)

hay `B<1/10`

4/3>1;1>3/4;4/3>3/4

1<11/9;9/11<11/9

100/99>1;1>99/100;100/99>99/100

minh nha cac ban 

4/3>1;1>3/4;4/3>3/4

1<11/9;9/11<11/9

100/99>1;1>9/100;100/99>99/100

4/3>1  ;  1>3/4  ;  4/3>3/4  ;  1<11/9  ;  9/11<11/9  ;  100/99>1  ;  1>99/100  ;  100/99>99/100

nha

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(B=\frac{1}{3}-\frac{1}{111}\)

\(B=\frac{12}{37}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(C=7.\frac{3}{35}\)

\(C=\frac{3}{5}\)

Ta có:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)

\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\) 

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(B=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}\)

\(\Rightarrow C=\frac{3}{5}\)

3^21*(1+3+3^2)+3^24*(1+3+3^2)+3^27*(1+3+3^2)=13*3^21+13*3^24+13*3^27=13*(3^21+3^24+3^27)chia hết cho 13

Giải nghĩa ^:mũ

                *:nhân

mk đang cần gấp người giải ai thấy bài này giải giúp mk nha 

Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)

\(=0\)

Do đó biểu thức trên đầu bài bằng 0

bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu

B= 3¹¹+3¹²+3¹³+...+3¹⁰¹

=>3B= 3¹²+3¹³+3¹⁴+...+3¹⁰¹

=>3B-B= 3¹²+3¹³+3^14+...+3¹⁰¹-3¹¹ -3¹²-3¹³-...-3¹⁰¹

=>2B=3¹⁰¹-3¹¹

=>B= 2¹⁰¹-3¹¹ :2