\(1,4.\dfrac{15}{49}-[\dfrac{4}{5}+\dfrac{2}{3}]:2\dfrac{1}{5}\)
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4a.
$A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}$
$2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}$
$\Rightarrow 2A-A=(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}})-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}})$
$\Rightarrow A=1-\frac{1}{2^{100}}$
4b.
$B=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}$
$3B=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}$
$\Rightarrow 3B-B=(3+1+\frac{1}{3}+...+\frac{1}{3^{99}})-(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}})$
$\Rightarrow 2B=3-\frac{1}{3^{100}}$
$\Rightarrow B=\frac{3}{2}-\frac{1}{2.3^{100}}$
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em ơi chưa có bài em nhé, em chưa tải bài lên lám sao mình giúp được
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`(3*x+2)^2=121`
\(=>\left[{}\begin{matrix}3x+2=11\\3x+2=-11\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=11-2\\3x=-11-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=9\\3x=-13\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{13}{3}\left(loại\right)\end{matrix}\right.\)
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\(A=1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\times4}+\dfrac{1}{4\times6}+\dfrac{1}{6\times8}+\dfrac{1}{8\times10}+\dfrac{1}{10\times12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{17}{12}\)
-5/21 nhé bn