b,-146+|-37|-(37-146)= -146 + 37 - 37+ 146

                                  = (-146+146)+(37-37)

                                  =0

a) (-14) + 5 + 17 + 14 = [(-14) + 14] + 5 + 17 = 22.

b) 40 +13 + (-25) + (-13) - 40 + (-25) + [l3 + (-13)] = 15.

c) (-5) + (-146) + (-15) +146 - (-5) + (-15)+[(-146) +146] = -20

d) (-2) + (-5) + 20 + (-13) = [(-2) + (-5) + (-13)] + 20 - 0

=\(\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)

=\(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

=\(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)

=\(\frac{15}{4}.\frac{2}{450}\)

=\(\frac{1}{60}\)

\(dungthikkothithoithanks\)

Thanh you

=>15x=146-71

=>15x=75

=>x=75:15

=>x=5

Vậy x=5

146 - 15 X = 71

         15 X = 146 - 71

         15 X = 75

              X = 75 : 15

              X = 5

146 - 15x = 71

=> 15x     = 146 - 71

=> 15x     = 75

=> x         = 75 : 15

=> x         = 5

Vậy x = 5

Câu 1: 

a:=17+5=22

b: =40-25=15

b: =>12-x=-7

hay x=19

$n_{Zn} = \dfrac{26}{65} = 0,4(mol)$
$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{HCl} = 2n_{H_2} = 0,4.2 = 0,8(mol)$
$a = \dfrac{0,8.36,5}{146}.100\% = 20\%$