\(\left(139,8-26,38\right):5,3+15,36\times4\\ =113,42:5,3+15,36\times4=21,4+61,44=82,84\)

\(82,84\)

( 139,8 - 26,38 ) : 5,3 + 15,36 x 4

= 113,42 : 5,3 + 15,36 x 4

= 21,4 + 15,36 x 4

= 21,4 + 61,44

= 82,84

a, 9,8-5,4-1,2=4,4-1,2=3,2
9,8-(5,4+1,2)=9,8-6,6=3,2
b,26,38-7,5-3,16=18,88-3,16=15,72
 26,38-(7,5+3,16)=26,38-10,66=15,72
c,37,86-9.2-4,8=28,66-4,8=23,86
37,86-(9,2+4,8)=37,86-14=23,86
 

139⁸₁₀ =139⁴₅

\(139,8=139\frac{4}{5}=\frac{699}{5}\)

\(5,3\div25\%+5,3\times7-5,3 \)

\(=5,3\div\dfrac{25}{100}+5,3\times7-5,3\)

\(=5,3\div\dfrac{1}{4}+5,3\times7-5,3\)

\(=5,3\times4+5,3\times7-5,3\times1\)

\(=5,3\times\left(7+4+1\right)\)

\(=5,3\times10\)

\(=53\)

   5,3: 25% + 5,3 \(\times\) 7 - 5,3

= 5,3 \(\times\) 4 + 5,3 \(\times\) 7 - 5,3 \(\times\) 1

= 5,3 \(\times\) ( 4 + 7 - 1)

= 5,3 \(\times\) ( 11 - 1)

= 5,3 \(\times\) 10

= 53 

sai rồi

Bảo toàn S:

\(\left(1+x\right)n_{oleum}=n_{H_2SO_4}=n_{BaSO_4}\)

\(n_{BaSO_4}=\dfrac{139,8}{233}=0,6mol\)

\(\Rightarrow n_{oleum}=\dfrac{0,6}{1+x}=\dfrac{51,6}{98+80x}\)

\(\Rightarrow x=2\)

Vậy oleum là \(H_2SO_4.3SO_3\)

= 5,3 x (13,5 - 2,5 - 1)

= 5,3 x 10

= 53 

2.8 là 2,8 nha

\(\left(5.3-2.8\right)-\left(4+5.3\right)\)

\(=2.5-4-5.3\)

=-6.8

=5,3-2,8-4-5,3

=(5,3-5,3)-2,8-4

=-6,8

=(5,3-5,3)+(5,1+4,9)=10

10

5,3-(-5,1)+(-5,3)+4,9

 =>(5,3-5,3)+(5,1+4,9)=10