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`1/5 . 4/7 + 3/7 . 1/5 -1/5`
`=1/5 . 4/7 + 3/7 . 1/5 -1/5 . 1`
`=1/5 . ( 4/7+3/7-1)`
`=1/5 . ( 7/7-1)`
`= 1/5 . 0`
`=0`
\(\dfrac{1}{5}\times\dfrac{4}{7}+\dfrac{3}{7}\times\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right)=\dfrac{1}{5}\times0=0\)
\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)
\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)
\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)
\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)
\(=-\dfrac{69}{45}\)
`6/7 . 8/13 +6/7 . 9/13+3/13 . 6/7`
`=6/7 . (8/13+9/13+3/13)`
`=6/7 . 20/13`
`=120/91`
\(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{7}.\dfrac{9}{13}+\dfrac{3}{13}.\dfrac{6}{7}\)
\(=\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}+\dfrac{3}{13}\right)\)
\(=\dfrac{6}{7}.\left(\dfrac{8+9+3}{13}\right)\)
\(=\dfrac{6}{7}.\dfrac{20}{13}\)
\(=\dfrac{6.20}{7.13}\)
\(=\dfrac{120}{91}\)
a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)
b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)
\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)
\(=\dfrac{20}{720}=\dfrac{1}{36}\)
`1/15+1/35+1/63+1/99+1/143`
`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`
`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`
`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`
`=1/2.(1/3-1/13)`
`=1/2 . 10/39`
`=5/39`
A= 1/3 + 1/3^2 + ... + 1/3^8
3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)
3A=1+ 1/3 + 1/3^2+ ... +1/3^7
=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)
=> 2A= 1 - 1/ 3^8
2A= 6560/6561
A= 6560/6561 : 2
A= 3280/6561
Theo đề, ta có: \(\dfrac{1+2x}{18}=\dfrac{1+4x}{34}\)
\(\Leftrightarrow34\left(1+2x\right)=18\left(1+4x\right)\)
\(\Leftrightarrow34+68x=18+72x\)
\(\Leftrightarrow34-18=72x-68x\)
\(\Leftrightarrow16=4x\)
\(\Leftrightarrow x=4\)
Khi \(x=4\) vào ta có: \(\dfrac{1+4.4}{34}=\dfrac{1+6.4}{2y^2}\Leftrightarrow\dfrac{1}{2}=\dfrac{25}{2y^2}\)
\(\Leftrightarrow2y^2=50\)
\(\Leftrightarrow y^2=50\)
\(\Leftrightarrow y=\pm5\)
\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\\ \dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\\ \Rightarrow x=-5\)
\(M=1+\dfrac{6}{2\cdot5}+\dfrac{10}{5\cdot10}+\dfrac{14}{10\cdot17}+\dfrac{18}{17\cdot26}\)
\(=1+2\left(\dfrac{3}{2\cdot5}+\dfrac{5}{5\cdot10}+\dfrac{7}{10\cdot17}+\dfrac{9}{17\cdot26}\right)\)
\(=1+2\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{26}\right)\)
\(=1+2\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=1+2\cdot\dfrac{12}{26}=1+\dfrac{24}{26}=\dfrac{50}{26}=\dfrac{25}{13}\)