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Ta có :
\(A=\frac{4}{1.5}+\frac{5}{5.10}+\frac{6}{10.16}+\frac{7}{16.23}+\frac{8}{23.31}+\frac{9}{31.40}\)
\(A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{31}+\frac{1}{31}-\frac{1}{40}\)
\(A=1-\frac{1}{40}=\frac{39}{40}\)
Ủng hộ mk nha !!! ^_^
Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)
=>\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}+\dfrac{1}{3^7}\)
=>\(3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^5}+\dfrac{1}{3^6}\)
=>\(3A-A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^5}+\dfrac{1}{3^6}-1-\dfrac{1}{3}-...-\dfrac{1}{3^7}\)
=>\(2A=3-\dfrac{1}{3^7}=\dfrac{3^8-1}{3^7}\)
=>\(A=\dfrac{3^8-1}{3^7\cdot2}\)
\(\frac{15}{57-x}=\frac{3}{8}\) <=> 3(57 - x) = 15*8 <=> 171 - 3x = 120 <=> 3x = 51 <=> x = 17
Vậy x = 17
\(\frac{15}{57-x}=\frac{3}{8}\Rightarrow5.8=57-x\Rightarrow x=57-40=17\)
Vậy x=17
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)
\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}+0+0+...+0\)
\(=\frac{7}{8}\)
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