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22 tháng 10 2017

\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{20}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\right)\)

\(A=1-\dfrac{1}{2^{20}}=\dfrac{2^{20}-1}{2^{20}}\)

Chọn A

20 tháng 12 2021

e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)

7 tháng 12 2021

\(B=1+\dfrac{1}{2}\cdot\dfrac{\left(1+2\right)\cdot2}{2}+\dfrac{1}{3}\cdot\dfrac{\left(1+3\right)\cdot3}{2}+...+\dfrac{1}{20}\cdot\dfrac{\left(20+1\right)\cdot20}{2}\\ B=1+\dfrac{3}{2}+2+\dfrac{5}{2}+...+10+\dfrac{21}{2}\\ B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{20}{2}+\dfrac{21}{2}\\ B=\dfrac{2+3+...+20+21}{2}=\dfrac{\dfrac{\left(21+2\right)\cdot20}{2}}{2}=\dfrac{23\cdot10}{2}=115\)

7 tháng 12 2021

em cảm ơn ạhihi

30 tháng 1 2023

\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)

\(=1+\dfrac{3\cdot2\div2}{2}+\dfrac{4\cdot3\div2}{3}+...+\dfrac{21\cdot20\div2}{20}\)

\(=1+\dfrac{3}{2}+2+...+\dfrac{21}{2}\) (A)

Trong (A) có \(\dfrac{\dfrac{21}{2}-1}{\dfrac{3}{2}-1}+1=20\) (số hạng)

Suy ra \(\left(A\right)=\left(\dfrac{21}{2}+1\right)\cdot20\div2=115\)

Vậy \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)=115\)

 

10 tháng 3 2017

\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)

\(\Rightarrow B=1+\dfrac{1}{2}.2.3\div2+\dfrac{1}{3}.3.4\div2+...+\dfrac{1}{20}.20.21\div2\)

\(\Rightarrow B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)

\(\Rightarrow B=\dfrac{2+3+4+...+21}{2}\)

\(\Rightarrow B=\dfrac{230}{2}\)

\(\Rightarrow B=115\)

Vậy \(B=115\)

10 tháng 12 2023

Sửa đề:

 \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)

PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)

=>\(x-1=\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)

27 tháng 11 2018

\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)

\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)

\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)

\(\Rightarrow11\left(x-2\right)=-20x+40\)

\(\Rightarrow11x-22=-20x+40\)

\(\Rightarrow11x+20x=22+40\)

\(\Rightarrow31x=62\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

20 tháng 12 2022

\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{23}{4}\)

\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=-\dfrac{77}{12}\)