tính
A= ( 1 + 1/3 ).(1+1/8).(1+1/15)...(1+1/24).(1+1/35)
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`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`-7/25 + (-8)/25`
`= (-7 - 8)/25`
`= -15/25`
`= -3/5`
`b)`
`6/13 + (-15)/39`
`= 18/39 + (-15)/39`
`= (18 - 15)/39`
`= 3/39`
`= 1/13`
`c)`
`5/7 + 4/(-14)`
`= 10/14 + (-4)/14`
`= (10 - 4)/14`
`= 6/14`
`= 3/7`
`d)`
`-8/18 + (-15)/27`
`= -4/9 + (-5)/9`
`= (-4-5)/9`
`= -9/9 = -1`
`2,`
`a)`
`3/5 + (-7)/4`
`= 12/20 + (-35)/20`
`= (12 - 35)/20`
`=-23/20`
`b)`
`(-2) + (-5)/8`
`= (-16)/8 + (-5)/8`
`= (-16 - 5)/8`
`= -21/8`
`c)`
`1/8 + (-5)/9`
`= 9/72 + (-40)/72`
`= (9-40)/72`
`= -31/72`
`d)`
`6/13 + (-14)/39`
`= 18/39 + (-14)/39`
`= (18 - 14)/39`
`= 4/39`
`e)`
`(-18)/24 + 15/21`
`= (-3)/4 + 5/7`
`= (-21)/28 + 20/28`
`= (-21 + 20)/28`
`= -1/28`
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
\(A=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)
\(=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(=\frac{1}{2}\times\left[\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\right)+\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}\right)+\left(\frac{4-2}{2\times4}+\frac{6-4}{4\times6}+\frac{8-6}{6\times8}+\frac{10-8}{8\times10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right]\)
\(=\frac{1}{2}\times\left[\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(=\frac{29}{45}\)
\(\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)
\(=\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)+\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}\right)\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)+\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
Bài 1:
a: \(=\dfrac{4}{8}+\dfrac{2}{8}+\dfrac{1}{8}=\dfrac{7}{8}\)
b: \(=\dfrac{4}{12}+\dfrac{2}{12}+\dfrac{1}{12}=\dfrac{7}{12}\)
B1:
a \(\dfrac{7}{8}\)
b \(\dfrac{7}{12}\)
B2:
a \(\dfrac{3}{7}\)
b \(\dfrac{5}{9}\)
c \(\dfrac{6}{5}\)
B3:
đội còn phải sửa phần quãng đường nữa :\(\dfrac{8}{15}\) phần quãng đường
=1/(1x3)x1/(2x4)x...x1/(9x11)
=1/(1x3x2x4x...x9x11)
=1/(1x2x3x3x4x4x5x5x...x9x10x11)
\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)
= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)
= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)
= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)
= \(\dfrac{4\times25\times36}{30\times24\times35}\)
= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)
Sai rồi nhé! Từ dấu bằng thứ 2 xuống dấu bằng thứ 3 bạn làm sao đc z?
2a= 2/3+2/8+2/15+2/24+2/35+2/48+2/63+2/80= [2/( 1*3)+2/( 3*5)+2/( 5*7)+2/( 7*9)]+[2/(2*4)+2/(4*6)+2/(6*8)+2/(8*10)]= [1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9]+[1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10]= [1/1-1/9]+[1/2-1/10]= 8/9+2/5= 58/45 =>a= 29/45
A= \(\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+\frac{1}{63}+\frac{1}{80}\)
A= \(\frac{2}{2}.\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\right)\)
A=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{9}+\frac{1}{2}-\frac{1}{10}\right)\)
A= tự tính
12/7
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