\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)

=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

                       \(=1-\frac{1}{2^9}=\frac{511}{512}\)

\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)

=>\(3B=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

=>\(3B-B=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}\right)\)

=>\(2B=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)

=>\(B=\frac{182}{243}:2=\frac{91}{243}\)

\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(A=1-\frac{1}{2^9}\)

\(A=\frac{2^9-1}{2^9}\)

Vậy \(A=\frac{2^9-1}{2^9}\)

Chúc bạn học tốt ~ 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)

A=511/512

B=91/243

A=511/512

B=91/243

A=511/512

B=91/243

\(3C=1+\frac{1}{3}+.....+\frac{1}{3^{2015}}\)

\(\Rightarrow3C-C=2C=\left(1+\frac{1}{3}+.....+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}.....+\frac{1}{3^{2015}}\right)=1-\frac{1}{3^{2015}}\)

\(\Rightarrow C=\frac{3^{2015}-1}{3^{2015}.2}\)

\(D=4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)

\(\Rightarrow3D=4\left(3+1+....+\frac{1}{3^4}\right)\)

\(\Rightarrow3D-D=2D=4\left(3+1+....+\frac{1}{3^4}\right)-4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)

\(\Rightarrow2D=4\left(3-\frac{1}{3^5}\right)\Rightarrow D=2\left(3-\frac{1}{3^5}\right)\)

các bn ơi giải giúp mình đi mà

a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)

       \(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)

       \(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\) 

       \(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)

       \(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)

       \(=3.\frac{196}{597}\)

       \(=\frac{196}{199}\)

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)