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Nhân 2 cả 2 vế lên:
\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243
\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)
\(24x+\frac{24}{25}=22x+\frac{224}{243}\)
\(2x=\frac{224}{243}-\frac{24}{25}\)
\(2x=-\frac{232}{6025}\)
\(x=\frac{-116}{6075}\)
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)
\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)
\(12x+\frac{12}{25}=11x+\frac{112}{243}\)
\(11x-12x=\frac{112}{243}-\frac{12}{25}\)
\(-1x=-\frac{116}{6075}\)
\(x=-\frac{116}{6075}\div\left(-1\right)\)
\(x=\frac{116}{6075}\)
Ta có: \(A=\frac{5}{18x23}+\frac{1}{23x24}+\frac{7}{24x31}+\frac{2}{31x33}+\frac{4}{33x37}+\frac{1}{37x38}+\frac{7}{38x45}+\frac{9}{45x54}\)
\(\Rightarrow A=\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{24}+\frac{1}{24}-\frac{1}{31}+...+\frac{1}{38}-\frac{1}{45}+\frac{1}{45}-\frac{1}{54}\)
\(\Rightarrow A=\frac{1}{18}-\frac{1}{54}\)
\(\Rightarrow A=\frac{3}{54}-\frac{1}{54}=\frac{1}{27}\)
Vậy \(A=\frac{1}{27}\)
*Công thức: \(\frac{a}{b.c}=\frac{1}{b}-\frac{1}{c}\left(a+b=c\right)\)
a) \(\frac{14}{15}+\frac{3}{11}+\frac{1}{15}+\frac{3}{11}+\frac{10}{22}\)
\(=\left(\frac{14}{15}+\frac{1}{15}\right)+\left(\frac{3}{11}+\frac{3}{11}\right)+\frac{10}{22}\)
\(=1+1+\frac{10}{22}\)
\(=2+\frac{10}{22}\)
\(=\frac{27}{11}\)
b) 9,25 + 9,25 x 8 + 9,25
= 9,25 x ( 8 + 1 + 1 )
= 9,25 x 10
= 92,5