Thực hiện phép tính:
a, 5/9 - ( 4/5 + 2/3 ) : 1 1/5
b, ( 3 3/8 : 5/6 - 6 5/8 : 5/6 ) . 5.6
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`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
a)-1-2-3-4-5-6-....-80
=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)
Khoảng cách giữa các số:(-1)-(-2)=1
Tổng các số hạng:(-1)-(-80)+1=80 số
Tổng:[(-1)+(-80)].80:2= -3240
=>-1-2-3-4-5-6+......-80=-3240
b,1-2+3-4+5-6+......+2021-2022
=(1-2)+(3-4)+(5-6)+...+(2021-2022)
=(-1)+(-1)+(-1)+...+(-1)
Tổng số cặp là:
(2022-1+1):2=1011 cặp
-1.1011=-1011
=>1-2+3-4+5-6+......+2021-2022= -1011
c, Đề bài sai
d,-4-8-12-16-.......-2020
=-4+(-8)+(-12)+(-16)+...+(-2020)
Khoảng cách giữa các số:-4-(-8)=4
Tổng các số hạng:[-4-(-2020]:4+1=505 số
Tổng:[-4+(-2020)].505:2=-511060
=>-4-8-12-16-.......-2020=-511060
a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)
\(=\dfrac{5}{7}\)
a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)
\(=8\cdot5^2+5-1\)
\(=200+4=204\)
b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)
\(=27-1+125-1\)
=150
a. (8,57 - 55 + 56) : 55
= (8,57 : 55) - (55 : 55) + (56 : 55)
= 1,72 - 1 + 5
= 2,89 - 1 + 5
= 6,89
b. (930 - 2719) : 357 + (1259 - 2512) : 524
= (930 : 357) - (2719 : 357) + (1259 : 524) - (2512 : 524)
= 33 - 1 + 125 - 1
= 27 - 1 + 125 - 1
= 150
c. (1012 + 511 . 29 - 513 - 28) : 4 . 55 . 106
= (1012 + 2,5 , 1010 - 513 - 28) : 1,25 . 1010
= (1012 : 1,25 . 1010) + (2,5 . 1010 : 1,25 . 1010) - (513 : 1,25 . 1010) - (28 : 1,25 . 1010)
= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)
= 81,90234373 \(\approx\) 82
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(=\dfrac{3}{5}+\dfrac{41}{25}\)
\(=\dfrac{15}{25}+\dfrac{41}{25}\)
\(=\dfrac{56}{25}\)
a) A = \(\dfrac{3}{5}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) \(\left(\dfrac{45}{4}-\dfrac{37}{4}\right)\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) . 2 : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{3}\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{25}\)
A = \(\dfrac{56}{25}\)
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)
\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)
b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)