Đặt \(A=1+3^2+3^4+...+3^{100}\)

\(9A=3^2+3^4+3^6+...+3^{102}\)

\(9A-A=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)

\(8A=3^{102}-1\)

\(A=\frac{3^{102}-1}{8}\)

Vậy \(A=\frac{3^{102}-1}{8}\)

Chúc bạn học tốt ~ 

Đặt A = 1 + 3^2 + 3^4 + 3^6 + ....+ 3^100

3^2A = 3^2 + 3^4 + 3^6 + ..+3^102

8A=3^2A - A = 3¹⁰² - 1

A = 3¹⁰² - 1/8

=. A = 3¹⁰² - 1 /8

\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)

`1/2+(-1/3)-(-5/4)`

`=1/2-1/3+5/4`

`=3/6-2/6+5/4`

`=1/6+5/4`

`=2/12+15/12`

`=17/12`

__

`5/4-1/2+(-7/8)`

`=5/4-1/2-7/8`

`=10/8-4/8-7/8`

`=6/8-7/8`

`=-1/8`

__

`1/5-1/2+9/10`

`=2/10-5/10+9/10`

`=-3/10+9/10`

`=6/10`

`=3/5`

__

`5/4-1/3+7/6`

`=15/12-4/12+14/12`

`=11/12+14/12`

`=25/12`

`#lv`

0,2x-2/3(x+1)=1/3

<=>0,2x-2/3x-2/3=1/3

<=>-7/15x=1

<=>x=-15/7

Vậy..............

Hok tốt

\(0,2x-\frac{2}{3}\left(x+1\right)=\frac{1}{3}\)

\(\frac{1}{5}x-\frac{2}{3}x+\frac{2}{3}=\frac{1}{3}\)

\(\left(\frac{1}{5}-\frac{2}{3}\right)x=-\frac{1}{3}\)

\(-\frac{7}{15}x=-\frac{1}{3}\)

\(x=\frac{5}{7}\)

=>A:1/2=1/1x3+1/3x5+1/5x7+...+1/99x101

=>2a=1/2(2/1x3+2/3x5+...+2/99x101)

từ đây tự làm

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(\Rightarrow2A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(\Rightarrow2A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(\Rightarrow4A=\frac{100}{101}\)

\(\Leftrightarrow A=\frac{100}{101}.\frac{1}{4}=\frac{4.25}{101.4}=25< 26\)

HS lớp 7 mà ko biết làm bài này người ta nói nó là thằng thiểu năng

a) |2x+1/3|=1/2

\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{3}=\frac{1}{2}\\2x+\frac{1}{3}=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{6}\\2x=\frac{-5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{12}\end{cases}}\)

b) |1-1/2x|=1/3

\(\Rightarrow\orbr{\begin{cases}1-\frac{1}{2}x=\frac{1}{3}\\1-\frac{1}{2}x=\frac{-1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{2}{3}\\\frac{1}{2}x=\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{8}{3}\end{cases}}\)

c) |3x+1|=1/5

\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{5}\\3x+1=\frac{-1}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{-4}{5}\\3x=\frac{-6}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{9}\\x=\frac{-2}{5}\end{cases}}\)

d) |x-1/2|+1=5/3

|x-1/2|=5/3-1

|x-1/2|=2/3

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{2}{3}\\x-\frac{1}{2}=\frac{-2}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{-1}{6}\end{cases}}}\)

a) \(3\dfrac{1}{4}=\dfrac{2}{3}:\left(\dfrac{-x}{2}\right)\Leftrightarrow\dfrac{13}{4}=\dfrac{2}{3}.\dfrac{-2}{x}\Leftrightarrow\dfrac{-2}{x}=\dfrac{39}{8}\Leftrightarrow x=-\dfrac{16}{39}\)

b) \(1-2\left(x+\dfrac{1}{3}\right)=\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|\Leftrightarrow1-2x-\dfrac{2}{3}=\dfrac{7}{15}\Leftrightarrow2x=-\dfrac{2}{15}\Leftrightarrow x=-\dfrac{1}{15}\)

c) \(\left(2x-1\right)\left(\dfrac{2}{5}-\dfrac{1}{3}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\dfrac{2}{5}-\dfrac{1}{3}x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{3}x=\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{6}{5}\end{matrix}\right.\)

d) \(-4\dfrac{3}{5}.2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\Leftrightarrow-10\le x\le-\dfrac{13}{7}\Leftrightarrow x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2;-1\right\}\)(do \(x\in Z\))

Bài 2: 

c: Ta có: \(\left(2x-1\right)\left(\dfrac{2}{5}-\dfrac{1}{3}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\dfrac{2}{5}-\dfrac{1}{3}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\\dfrac{1}{3}x=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{6}{5}\end{matrix}\right.\)

Vì tích A có lẻ thừa số âm là 2011 => Tích A mang dấu âm

Mà số hạng của tích này đều có cơ số bằng ( - 1 )

=> A = - 1