\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)

=>\(B=\dfrac{32+16+6+2+1}{64}\)

=>\(B=\dfrac{63}{64}\)

\(\dfrac{63}{64}\)

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)

\(2B=1+\dfrac{1}{2}+...+\dfrac{1}{2^5}\)

\(\Rightarrow2B-B=1-\dfrac{1}{2^6}\)

\(\Rightarrow B=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)

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Lời giải:

\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)

\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)

\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)

\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)

\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)

\(x=\dfrac{35.16}{-28}\)

\(x=-20\)

\(\dfrac{-28}{35}=\dfrac{16}{x}\\ \Rightarrow-28\cdot x=35\cdot16=560\\ x=560:-28\\ x=-20\)

\(x=\dfrac{35.16}{-28}=20\)

2.(-x-3)=53-27

2.(-x-3)=26

-x-3=26/2

-x-3=13

-x=13+3

-x=16

=> x=-16

2.(-x-3)+27=53

2.(-x-3)      =53-27

2.(-x-3)      =26

   (-x-3)      =26:2

   -x-3      =13

   => -x=  16

  => x= -16

2.(-x-3)+27=53

2.(-x-3)=53-27

2.(-x-3)=26

-x-3=26:2

-x-3=13

-x=13+3

-x=16

=>x=-16

Vậy x=-16

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\) \(\Rightarrow A< \dfrac{99}{100}\)

\(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\right)=1-A>\dfrac{1}{100}\)

N=1/2+1/2²+...+1/2¹⁰

2N=1+1/2+...+1/2⁹

2N-N=1-1/2¹⁰=1-1/1024=1023/1024

Giải:

N=1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰

2N=1+1/2+1/2²+...+1/2⁸+1/2⁹

2N-N=(1+1/2+1/2²+...+1/2⁸+1/2⁹)-(1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰)

N=1-1/2¹⁰=1023/1024

Chúc bạn học tốt!

\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)

\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)