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`@` `\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) 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: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
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, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)
\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)
\(\frac{6}{5}-\frac{1}{4}+\frac{4}{5}-\frac{3}{4}\)
\(=\left(\frac{6}{5}+\frac{4}{5}\right)-\left(\frac{1}{4}+\frac{3}{4}\right)\)
\(=2-1\)
\(=1\)
\(\frac{7}{9}-\frac{5}{12}+\frac{2}{9}-\frac{7}{12}\)
\(=\left(\frac{7}{9}+\frac{2}{9}\right)-\left(\frac{5}{12}+\frac{7}{12}\right)\)
\(=1-1\)
\(=0\)
các câu sau tương tự
A=1
B=0
C=ÂM 8/15
D=0
E=2
F=3/2
H=2/3
LẦN SAU CHO KHÓ SÍU NHA LỚP 5 CŨNG LÀM ĐC
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
i, \(\dfrac{1}{7}\) .(- \(\dfrac{3}{8}\)) + \(\dfrac{1}{7}\).(-\(\dfrac{13}{8}\))
=- \(\dfrac{1}{7}\).( \(\dfrac{3}{8}\)+ \(\dfrac{13}{8}\))
= - \(\dfrac{1}{7}\). 2
= -2/7
k, -4\(\dfrac{1}{3}\). ( \(\dfrac{1}{2}\)- \(\dfrac{1}{6}\))
= - \(\dfrac{13}{3}\). \(\dfrac{1}{3}\)
= - 13/9
l, \(\dfrac{3}{5}\).\(\dfrac{13}{46}\)- \(\dfrac{1}{10}\).\(\dfrac{16}{23}\)
= \(\dfrac{39}{230}\) - \(\dfrac{16}{230}\)
= \(\dfrac{1}{10}\)
m, (- \(\dfrac{3}{4}\)+ \(\dfrac{2}{5}\)): \(\dfrac{3}{7}\) + ( \(\dfrac{3}{5}\)- \(\dfrac{1}{4}\)): \(\dfrac{3}{7}\)
= (- \(\dfrac{3}{4}\) + \(\dfrac{2}{5}\)+ \(\dfrac{3}{5}\)- \(\dfrac{1}{4}\)): \(\dfrac{3}{7}\)
= 0 : \(\dfrac{3}{7}\)
= 0
t, \(\dfrac{7}{8}\): ( \(\dfrac{2}{9}\) - \(\dfrac{1}{18}\)) + \(\dfrac{7}{8}\): ( \(\dfrac{1}{36}\) - \(\dfrac{5}{12}\))
= \(\dfrac{7}{8}\): \(\dfrac{1}{6}\) + \(\dfrac{7}{8}\): (- \(\dfrac{7}{18}\))
= \(\dfrac{21}{4}\) - \(\dfrac{9}{4}\)
= 3
= \(\dfrac{12}{4}\)