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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 , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)
= \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)
b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)
= \(\frac{1}{5}+0=\frac{1}{5}\)
c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)
= \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)
= \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)
= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)
\(\frac{6}{5}-\frac{1}{4}+\frac{4}{5}-\frac{3}{4}\)
\(=\left(\frac{6}{5}+\frac{4}{5}\right)-\left(\frac{1}{4}+\frac{3}{4}\right)\)
\(=2-1\)
\(=1\)
\(\frac{7}{9}-\frac{5}{12}+\frac{2}{9}-\frac{7}{12}\)
\(=\left(\frac{7}{9}+\frac{2}{9}\right)-\left(\frac{5}{12}+\frac{7}{12}\right)\)
\(=1-1\)
\(=0\)
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
Các bài toán này bn bấm máy thì sẽ ra nha.
Bài 7 có quy tắc tính trong ngoặc trước, ngoài ngoặc sau.
7b bn nhóm 2/3 +1/18 lại vào một chỗ và để cho chúng dấu ngoặc
tíc mình nha