A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
 

 C = 2^-3 + (5²)³.5^-3 + 4^-3.16 - 2.3² - 105.(\(\dfrac{24}{51}\))⁰

C =  \(\dfrac{1}{8}\) + 5⁶.5^-3 + 4^-3.4² - 2.9 - 105.1

C =  \(\dfrac{1}{8}\) + 5³ + \(\dfrac{1}{4}\) - 18 - 105

C =  (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\))  - (105 - 125 + 18)

C = \(\dfrac{3}{8}\) - (-20 + 18)

C = \(\dfrac{3}{8}\)  + 2

C = \(\dfrac{19}{8}\)

C1: \(A=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{150}{30}+\frac{50}{30}-\frac{45}{30}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)

\(=\frac{35}{6}-\frac{155}{30}-\frac{19}{6}=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{15}{6}=-2\frac{1}{2}\)

C2: \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)

\(=-2-0-\frac{1}{2}=-2\frac{1}{2}\)

a) (-37) + 14 + 26 + 37

= [(-37) + 37] + (14 + 26)

= 0 + 40 = 40

b) (-24) + 6 + 10 + 24

= [(-24) + 24] + (10 + 6)

= 0 + 16 = 16

c) 15 + 23 + (-25) + (-23)

= [15 + (-25)] + [23 + (-23)]

= (-10) + 0 = -10

d) 60 + 33 + (-50) + (-33)

= [60 + (-50)] + [33 + (-33)]

= 10 + 0 = 10

e) (-16) + (-209) + (-14) + 209

= [(-16)  + (-14)] + [(-209) + 209]

= (-30) + 0 = -30

f) \(-3^2+\left(-54\right)\div\left[\left(-2\right)^8+7\right]\times\left(-2\right)^2\\ =\left(-9\right)+\left(-54\right)\div263\times4\\ =\left(-9\right)+\dfrac{-216}{263}=\dfrac{-2583}{263}\)

a. \(\left[\left(-37\right)+37\right]+\left(14+16\right)\) = 30 

B. \(\left[\left(-24\right)+24\right]+\left(10+6\right)\) = 16

C. \(\left[\left(-23\right)+23\right]+\left(15-23\right)\)= -8

d. \(\left[33-33\right]+\left(60-50\right)\) = 10

e. \(\left(209-209\right)+\left(-16-14\right)\)= -30