\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)

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

\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4E=3-\frac{203}{3^{100}}< 3\)

=> 4E < 3 => E < 3/4 

bài 1 mifk viết sai nha.

bài 1: cho A=1+3+3\(^2\)+3\(^3\)+...+3\(^{10}\).Tìm số tự nhiên n biết 2 x A + 1 = 3\(^n\)

B1:

\(A=1+3+3^2+3^3+...+3^{10}\\ 3A=3+3^2+3^3+3^4+...+3^{11}\\ 3A-A=3^{11}-1\\ \Rightarrow A=\frac{3^{11}-1}{2}\)

mấy câu khác tương tự nha

a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)

= \(\dfrac{-7}{9}.\dfrac{11}{4}\)

= \(\dfrac{-77}{36}\)

b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)

= \(\dfrac{2}{3}+\dfrac{-2}{15}\)

= \(\dfrac{10}{15}+\dfrac{-2}{15}\)

= \(\dfrac{-8}{15}\)

c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)

= \(\dfrac{2}{3}-4.\dfrac{5}{4}\)

= \(\dfrac{2}{3}-5\)

= \(\dfrac{-13}{3}\)

d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)

= \(\dfrac{1}{2}\) . 11 - 7

= \(\dfrac{11}{2}-\dfrac{14}{2}\)

= \(\dfrac{-3}{2}\)

e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)

= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)

= \(\dfrac{3}{4}.-28\)

= \(-21\)

1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3

=>x=-16/3:7/3=-7/16

2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35

=>|x-2|=129/35

=>x-2=129/35 hoặc x-2=-129/35

=>x=199/35 hoặc x=-59/35

các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho

\(E=\frac{1}{3}+\frac{2}{3}+\frac{3}{3}+...+\frac{100}{3}\)

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

\(=\frac{101\times100\div2}{3}\)

\(=\frac{5050}{3}\)

 vì \(\frac{5050}{3}>1\)mà \(\frac{3}{4}< 1\)\(\Rightarrow\frac{5050}{3}>\frac{3}{4}\)

 Vậy E>\(\frac{3}{4}\)có thể bạn ghi sai đề phải là e>3/4 mới đúng

bạn ơi có mũ đấy tử là bao nhieu thì mũ của mẫu là từng đây 

a; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2

1 = 1 = 1

1/2^2 < 1/1.2 = 1/1 - 1/2

1/3^2 < 1/2.3 = 1/2 - 1/3

.......................

1/100^2 < 1/99.100 = 1/99 - 1/100

Cộng vế với vế ta có:

A = 1 + 1 - 1/100

A = 2 - 1/100 < 2 (đpcm)