(-12.\(\dfrac{2}{7}\) + \(\dfrac{8}{9}\): 3\(\dfrac{1}{2}\) - \(\dfrac{2}{7}\).\(\dfrac{5}{18}\)). 3\(\dfrac{1}{2}\)

(-\(\dfrac{24}{7}\) +  \(\dfrac{8}{9}\):\(\dfrac{7}{2}\) - \(\dfrac{5}{63}\)). \(\dfrac{7}{2}\)

= (-\(\dfrac{24}{7}\) + \(\dfrac{16}{63}\) - \(\dfrac{5}{63}\)).\(\dfrac{7}{2}\)

= (\(-\dfrac{216}{63}\) + \(\dfrac{16}{63}\) - \(\dfrac{5}{63}\)). \(\dfrac{7}{2}\)

=  - \(\dfrac{205}{63}\). \(\dfrac{7}{2}\)

= - \(\dfrac{205}{18}\) 

a: ta có: \(5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5\)

\(=5\cdot\left(2^3\right)^{12}\cdot\left(3^2\right)^5\cdot3^2\cdot\left(-1\right)\cdot2^{20}\)

\(=-5\cdot3^{12}\cdot2^{56}\)

Ta có: \(7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5\)

\(=-7\cdot\left(7^2\right)^3\cdot\left(3^4\right)^5\cdot7^{10}\cdot\left(2^3\right)^5\)

\(=-7\cdot7^6\cdot3^{20}\cdot7^{10}\cdot2^{15}=-7^{17}\cdot2^{15}\cdot3^{20}\)

Ta có: \(B=\frac{5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5}{7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5}\)

\(=\frac{-5\cdot3^{12}\cdot2^{56}}{-7^{17}\cdot2^{15}\cdot3^{20}}=\frac{5\cdot2^{41}}{7^{17}\cdot3^8}\)

b: \(13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4\)

\(=13\cdot2^{14}\cdot3^{20}-3^{14}\cdot2^{16}\)

\(=3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)\)

\(15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}\)

\(=3\cdot5\cdot3^{12}\cdot2^8+3\cdot\left(3^3\right)^5\cdot\left(2^3\right)^{10}=3^{13}\cdot5\cdot2^8+3^{16}\cdot2^{30}\)

\(=3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)\)
Ta có: \(C=\frac{13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4}{15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}}\)

\(=\frac{3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)}{3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)}=\frac{3\cdot2^6\cdot\left(13\cdot3^6-2^2\right)}{5+3^3\cdot2^{22}}\)

a) 2/5 < x < 6/5

=> x = 1 ( =5/5 )    (vì x thuộc Z)

Vậy x = 1

b) 3/5 < 3/x < 3/2

=> 5 > x > 2

=> x thuộc { 4 ; 3 }   (vì x thuộc Z)

Vậy ...

c) 3/8 + -11/8 < x < 22/9 + 5/18

=> -8/8 < x < 49/18

=>-1 < x < 2+13/18

=> x thuộc {0; 1; 2}  ( vì x thuộc Z )

Vậy...

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(2Fe+O_2\underrightarrow{t^o}2FeO\)

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)

\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)

\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)

\(=\dfrac{40+\left(-135\right)+30}{180}\)

\(=\dfrac{-65}{180}\)

\(=\dfrac{-13}{36}\)

2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)

\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)

\(=\dfrac{-21}{36}-\dfrac{22}{36}\)

\(=\dfrac{-21-22}{36}\)

\(=\dfrac{-43}{36}\)

3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)

\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)

\(=\dfrac{14}{16}-\dfrac{-5}{16}\)

\(=\dfrac{14-\left(-5\right)}{16}\)

\(=\dfrac{19}{16}\)

4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)

\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)

\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)

\(=\dfrac{30-\left(-72\right)-25}{80}\)

\(=\dfrac{77}{80}\)

1)` 2/9 + -3/4 + 5/30= (40+(-135) + 30 )/180 = -65/180 = -13/36`

2) `-7/12 - 11/18=(-21)/36 - 22/36 = -43/36 `

3) `7/8 - (-5)/16=14/16- (-5)/16 = 19/16 `

4) `3/8 - (-9)/10 - 5/16= (30- (-72) -25)/80=77/80`

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17+18 = 171 nhé

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17+18.  = 171

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

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