\(E=1-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{10}}\)

\(E=1-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)

Đặt \(S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\)

\(2S=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)

\(2S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

\(2S-S=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(S=\dfrac{1}{2}-\dfrac{1}{2^{10}}\). Khi đó \(E=1-\left(\dfrac{1}{2}-\dfrac{1}{2^{10}}\right)=1-\dfrac{1}{2}+\dfrac{1}{2^{10}}=\dfrac{513}{1024}\)

2E=2-1-\(\frac{1}{2^2}\)-\(\frac{1}{2^3}\)-.....-\(\frac{1}{2^{11}}\)

2E+E=2-\(\frac{1}{2^{11}}\)

3E=2-\(\frac{1}{2^{11}}\)

E=2-\(\frac{1}{2^{11}}\):3

\(E=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)

\(E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(F=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+...+\frac{15}{146\cdot150}\)

\(F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

\(\Rightarrow F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)

\(G=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(G=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(G=\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+\frac{5}{10\cdot13}+...+\frac{5}{25\cdot28}\)

\(G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(\Rightarrow G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)

sao nhiều vậy bạn 

a) Sai đề

b)\(\left(x-2\right)^6=4^3\)

\(\left(x-2\right)^6=4^3=64=2^6=\left(-2\right)^6\)

\(\Rightarrow x-2=2\)hoặc \(x-2=-2\)

\(Th1:x-2=2\Leftrightarrow x=4\)

\(Th2:x-2=-2\Leftrightarrow x=0\)

c)\(\left(x+1\right)^2=3^4\)

\(\left(x+1\right)^2=3^4=81=9^2=\left(-9\right)^2\)

\(\Rightarrow x+1=9\)hoặc \(x+1=-9\)

\(Th1:x+1=9\Leftrightarrow x=8\)

\(Th2:x+1=-9\Leftrightarrow x=-10\)

d) \(3^{x+1}=243\)

\(3^{x+1}=3^5\)

\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)

e)\(2^x=4^{10}=\left(2^2\right)^{10}=2^{20}\Leftrightarrow x=20\)

d) \(3^{x+1}=243\)

\(3^{x+1}=3^5\)

\(x=5-1\)

\(x=4\)

e) \(2^x=4^{10}\)

\(2^x=\left(2^2\right)^{10}\)

\(\Rightarrow2^x=2^{20}\)

\(\Rightarrow x=20\)

a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)

hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)

Vậy: \(x=\frac{9}{10}\)

b) Ta có: \(5\frac{4}{7}:x=13\)

\(\Leftrightarrow\frac{39}{7}:x=13\)

\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)

Vậy: \(x=\frac{3}{7}\)

c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=84\)

\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)

Vậy: x=30

d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)

hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)

Vậy: x=-5

e) Ta có: \(8\frac{2}{3}:x-10=-8\)

\(\Leftrightarrow\frac{26}{3}:x=2\)

hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)

Vậy: \(x=\frac{13}{3}\)

g) Ta có: \(x+30\%=-1.3\)

\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)

hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)

Vậy: \(x=\frac{-8}{5}\)

i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)

\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)

\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)

\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)

Vậy: x=-9

k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)

hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)

Vậy: x=30

m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)

\(\Leftrightarrow\left|2x-1\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

thank you nha!