Question

[Nguyễn Tuấn Phi]

Answer 1

[Akai Haruma]

a. 

\(=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}+\frac{-41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)

b.

\(=\frac{-20}{24}+\frac{9}{24}-\frac{1}{10}=\frac{-11}{24}+\frac{1}{10}=\frac{-43}{120}\)

c.

\(=\frac{-25}{27}-\frac{31}{42}+\frac{7}{27}+\frac{3}{42}\\ =\frac{-25}{27}+\frac{7}{27}+(\frac{-31}{42}+\frac{3}{42})\\ =\frac{-2}{3}+\frac{-2}{3}=\frac{-4}{3}\)

Answer 2

[Akai Haruma]

d.

\(=\frac{5}{36}+\frac{-28}{36}-\frac{5}{4}=\frac{5}{36}+\frac{-28}{36}-\frac{45}{36}\\ =\frac{-68}{36}=\frac{-17}{9}\)

e. 

\(=\frac{3}{5}+\frac{2}{5}: 2=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}\)

f.

\(=\frac{3}{10}(\frac{-23}{7}+\frac{13}{7})=\frac{3}{10}.\frac{-10}{7}=\frac{-3}{7}\)

g.

\(=\frac{3}{2}+\frac{1}{4}+(-1)=\frac{7}{4}-1=\frac{3}{4}\)

h.

\(=\frac{2^{2023}}{(2^2)^{1011}}=\frac{2^{2023}}{2^{2022}}=2^{2023-2022}=2\)

i.

\(=4-4+(-8).\frac{5}{4}=0+(-10)=-10\)