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28 tháng 7 2023

\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)

\(=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{7}\right)+...+\left(\dfrac{1}{512}+\dfrac{1}{513}+...+\dfrac{1}{1023}\right)+\dfrac{1}{1024}< 1+\dfrac{1}{2}.2+\dfrac{1}{2^2}.2^2+...+\dfrac{1}{2^9}.2^9+\dfrac{1}{1024}\)

\(=1+1+1+1+...+\dfrac{1}{1024}=10+\dfrac{1}{1024}< 11\left(đpcm\right)\)

27 tháng 7 2023

Bạn xem lại đề

27 tháng 7 2023

đề đúng mà bạn

 

\(\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{4}\right)^5\cdot3}{\dfrac{1}{1024}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)

\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot2}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)

\(=2:\dfrac{-1}{6}=2\cdot\left(-6\right)=-12\)

3 tháng 2 2017

\(\Rightarrow B=\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{2}\right)}\) \(\Rightarrow B=\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\) \(\Rightarrow B=\frac{\left(\frac{1}{2}\right)^{10}.2}{\left(\frac{1}{2}\right)^{10}.\left(-\frac{1}{6}\right)}=-12\)

29 tháng 9 2015

B = \(\frac{\frac{1}{2^{10}}.5-\frac{1}{\left(2^2\right)^5}.3}{\frac{1}{2^{10}}.\frac{1}{3}-\frac{1}{2^{11}}}=\frac{\frac{1}{2^{10}}.\left(5-3\right)}{\frac{1}{2^{10}}.\left(\frac{1}{3}-\frac{1}{2}\right)}=\frac{2}{\left(-\frac{1}{6}\right)}=2:\left(-\frac{1}{6}\right)=-12\)

31 tháng 1 2019

\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)

31 tháng 1 2019

\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)

4 tháng 2 2019

1) tự làm (thực hiện từ dưới lên)

2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)

      = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)

     = \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)

     = \(\frac{2}{-\frac{1}{6}}\)= 2 . (-6) = -12

4 tháng 2 2019

1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)

24 tháng 9 2017

\(a.\)

\(1-\dfrac{1}{2}\left(\dfrac{3}{2}-2x\right)=4x-\dfrac{1}{4}\)

\(\Rightarrow1-\dfrac{3}{4}+x=4x-\dfrac{1}{4}\)

\(\Rightarrow1-\dfrac{3}{4}+\dfrac{1}{4}=4x-x\)

\(\Rightarrow3x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{6}\)

\(b.\)

\(x^{10}=1024\)

\(\Rightarrow x^{10}=2^{10}\)

\(\Rightarrow x=2\)

\(c.\)

\(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)