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a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
`@` `\text {Ans}`
`\downarrow`
`1)`
`5/7*37 13/23 - 51 13/23*5/7`
`= 5/7* (37 13/23 - 51 13/23)`
`= 5/7* (-14)`
`= -10`
`2)`
`-2/3 +1/3+0,5+2 1/2`
`= -2/3 + 1/3 + 1/2 + 5/2`
`= (-2/3+1/3) + (1/2+5/2)`
`= -1/3 + 3`
`=8/3`
`3)`
`-0,5+2/3+1/2`
`= -1/2 + 2/3 + 1/2`
`= (-1/2 + 1/2) + 2/3`
`= 2/3`
`4)`
`(8+2 1/3-3/5) -(5+0,4)-(3 1/2 -2)`
`= 8+ 7/3 - 3/5 - 5 - 0,4 - 7/2 + 2`
`= (8+2-5) + (-3/5 - 2/5) + (7/3 - 7/2)`
`= 5 - 1 - 7/6`
`= 4 - 7/6 = 17/6`
`5)`
`(2/9-7/12):3/4+(16/9-5/12):3/4`
`= (2/9 - 7/12) \times 4/3 + (16/9 - 5/12) \times 4/3`
`= 4/3 *(2/9 - 7/12 + 16/9 - 5/12)`
`= 4/3 * [(2/9 + 16/9) + (-7/12 - 5/12)]`
`= 4/3 * ( 2 - 1)`
`= 4/3 * 1 = 4/3`
`6)`
`-(2021.0,7+19,75) +0,7- (8-19,75)`
`= -2021*0,7 -19,75 + 0,7 - 8 + 19,75`
`= 0,7*(-2021 + 1) - 8`
`= -1414-8`
`= -1422`
`7)`
`15/34+7/21+19/34-20/15`
`= (15/34 + 19/34) + 7/21 - 20/15`
`= 1 + 7/21 - 20/15`
`= 4/3 - 20/15 =0`
`8)`
`2 5/6+1/6:(-5/8)`
`= 17/6 + (-4/15)`
`= 77/30`
`9)`
`(-2)^2 +2/9. (4/5-2/3)`
`= 4 + 2/9*2/15`
`= 4+4/135`
`= 544/135`
`10)`
`(-1/5+3/7):5/4+(-4/5+4/7):5/4`
`= (-1/5+3/7) * 4/5 + (-4/5+4/7) * 4/5`
`= 4/5*(-1/5 +3/7-4/5+4/7)`
`= 4/5*[(-1/5-4/5)+(3/7+4/7)]`
`= 4/5* (-1+1)`
`= 4/5*0=0`
`11)`
`2022,2021 . 1954,1945+ 2022,2021 . (-1954,1945)`
`= 2022,2021 * [1954,1945 + (-1954,1945)]`
`= 2022,2021*0 `
`= 0`
`12)`
`-5,2 .72 +69,1 +5,2 . (-28)+(-1,1)`
`= -5,2*72 + 69,1 - 5,2*28 - 1,1`
`= -5,2*(72+28) + (69,1 - 1,1)`
`= -5,2*100 + 68`
`= -520 + 68`
`= -452`
`13)`
`(7 -1/2-3/4) : (5-1/4-5/8)`
`= 23/4 \div 33/8`
`=46/33`
`14)`
`(8+ 2 1/3 -3/5) -(5+0,4) -( 3 1/3 - 2)`
`= 8+ 2 1/3 - 3/5 - 5 - 0,4 - 3 1/3 + 2`
`= (8+2-5) + (2 1/3 - 3 1/3) - (0,6 + 0,4) `
`= 5 - 1 - 1`
`= 3`
a, \(\frac{-3}{7}+\frac{5}{13}-\frac{4}{7}+\frac{8}{13}\)
\(=\frac{-3}{7}-\frac{4}{7}+\frac{5}{13}+\frac{8}{13}\)
\(=-\frac{7}{7}+\frac{13}{13}=-1+1=0\)
b, \(\frac{-5}{14}-\frac{2}{-14}+\frac{1}{8}+\frac{1}{8}\)
\(=\frac{-5}{14}+\frac{2}{14}+\frac{1}{8}+\frac{1}{8}\)
\(=-\frac{3}{14}+\frac{1}{4}=\frac{1}{28}\)
c,\(-\frac{5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\)
\(=-\frac{5}{13}-\frac{3}{13}-\frac{3}{5}+\frac{4}{10}\)
\(=-\frac{8}{13}-\frac{3}{5}+\frac{4}{10}=-\frac{79}{65}+\frac{4}{10}=-\frac{53}{65}\)
d, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)
\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)
Bài 2:
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{100}{101}\)
Vậy ...
\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}.\dfrac{15}{16}\)
\(\Leftrightarrow B=\dfrac{5}{16}\)
Vậy ...
Bài 1:
B=\(\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}\right)}\)
\(=\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}\)
\(=\dfrac{\left(\dfrac{2^4+2^3+2^2+2+1}{2^4}\right)}{\left(2^4-2^3+2^2-2+1\right)}\)
\(=\dfrac{\left(2^3+2\right)\left(2+1\right)+1}{2^4}.\dfrac{2^4}{\left(2^3+2\right)\left(2-1\right)}\)
\(=\dfrac{2\left(2^2+1\right)\left(2+1\right)+1}{2\left(2^2+1\right)\left(2-1\right)+1}\)
\(=\dfrac{2.5.3+1}{2.5.1+1}\)
\(=\dfrac{31}{11}\)
\(=2,\left(81\right)\)