1) 5 + (-4) = 1

2) (-8) + 2 = -6

3) 8 + (-2) = 6

4) 11 + (-3) = 8

5) (-11) + 2 = -9

6) (-7) + 3 = -4

7) (-5) + 5 = 0

8) 11 + (-12) = -1

9) (-18) + 20 = 2

10) (15) + (-12) = 3

11) (-17) + 17 = 0

12) 16 + (-2) = 14

13) (30) + (-14) = 16

14) (-19) + 20 = 1

15) (-18) + 15 = -3

16) (10) + (-6) = 4

17) (-28) + 14 = -14

18) 15 + (-30) = -15

19) (15) + (-4) = 11

20) (-21) + 11 = -10

21) 8 + (-22) = -14

22) (-15) + 4 = -11

23) (-3) + 2 = -1

24) 17 + (-14) = 3

25) 17 + (-14) = 3

thế này đọc đc mới đỉnh

a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)

b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)

c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)

d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)

bai EZ​ qua

mà bài dài quá sao làm hết dc

`@` `\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.`

\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)

\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)

\(=0,2-\dfrac{2}{3}\)

\(=-\dfrac{7}{15}\)

\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)

\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)

\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

#\(Toru\)

\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)

1: \(=-\dfrac{3}{4}+3-\dfrac{1}{4}-\dfrac{5}{4}+\dfrac{9}{2}=-\dfrac{9}{4}+3+\dfrac{9}{2}=\dfrac{-9+12+18}{4}=\dfrac{21}{4}\)

2: \(=\dfrac{-5}{7}\left(\dfrac{13}{19}+\dfrac{6}{19}\right)+\dfrac{5}{7}=\dfrac{-5}{7}+\dfrac{5}{7}=0\)

3: \(=\dfrac{2}{5}\left(1-\dfrac{8}{3}-\dfrac{5}{8}\right)=\dfrac{2}{5}\cdot\dfrac{-55}{24}=\dfrac{-11}{12}\)