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16 tháng 4 2022

x=1/16

21 tháng 12 2021

giúp mình đi ạ khocroi

 

21 tháng 12 2021

\(=\dfrac{1}{x+1}-\dfrac{8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-4-8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-12}{\left(x+1\right)\left(x-4\right)}=\dfrac{-11}{2\cdot\left(-3\right)}=\dfrac{11}{6}\)

DD
3 tháng 6 2021

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)

\(A=1-\frac{1}{128}\)

\(A=\frac{127}{128}\)

DD
3 tháng 6 2021

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

\(B=1-\frac{1}{16}=\frac{15}{16}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow4\times x+\frac{15}{16}=1\)

\(\Leftrightarrow4\times x=\frac{1}{16}\)

\(\Leftrightarrow x=\frac{1}{64}\)

13 tháng 6 2021

Hơi căng 

15 tháng 1

Ta có: \(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{x+1-\left(x-1\right)}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{2}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{2\left(x^2+1\right)-2\left(x^2-1\right)}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2x^2+2-2x^2+2}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4\left(x^4+1\right)-4\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4x^4+4-4x^4+4}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{8}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}=\frac{8x^8+8-8x^8+8}{x^{16}-1}-\frac{16}{x^{16}+1}\)

\(=\frac{16}{x^{16}-1}-\frac{16}{x^{16}+1}=\frac{16x^{16}+16-16x^{16}+16}{\left(x^{16}-1\right)\left(x^{16}+1\right)}=\frac{32}{x^{32}-1}\)

12 tháng 11 2025

a:Sửa đề: 1x3+2x4+...+99x101

\(=1\times\left(1+2\right)+2\times\left(2+2\right)+\cdots+99\times\left(99+2\right)\)

\(=\left(1\times1+2\times2+\cdots+99\times99\right)+2\times\left(1+2+\cdots+99\right)\)

\(=\frac{99\times\left(99+1\right)\times\left(2\times99+1\right)}{6}+2\times\frac{99\times100}{2}\)

\(=\frac{99\times100\times199}{6}+99\times100=33\times50\times199+99\times100\)

\(=33\times50\times\left(199+3\times2\right)=33\times50\times205=338250\)

b: \(\frac89\times\frac{15}{16}\times\ldots\times\frac{2499}{2500}\)

\(=\left(1-\frac19\right)\times\left(1-\frac{1}{16}\right)\times\ldots\times\left(1-\frac{1}{2500}\right)\)

\(=\left(1-\frac13\right)\times\left(1-\frac14\right)\times\ldots\times\left(1-\frac{1}{50}\right)\times\left(1+\frac13\right)\times\left(1+\frac14\right)\times\ldots\times\left(1+\frac{1}{50}\right)\)

\(=\frac23\times\frac34\times\ldots\times\frac{49}{50}\times\frac43\times\frac54\times\ldots\times\frac{51}{50}=\frac{2}{50}\times\frac{51}{3}=\frac{17}{25}\)

1 tháng 3 2022

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