Ta có: \(2\frac{1}{3}+\frac{11}{5}:33-\frac{1}{50}.\left(-5\right)^2\)

\(=\frac{7}{3}+\frac{11}{5}.\frac{1}{33}-\frac{1}{50}.25\)

\(=\frac{7}{3}+\frac{11}{165}-\frac{25}{50}\)

\(=\frac{7}{3}+\frac{1}{15}-\frac{1}{2}\)

\(=\frac{70}{30}+\frac{2}{30}-\frac{15}{30}\)

\(=\frac{56}{30}=\frac{28}{15}\)

Câu 2:

a: \(\frac{5}{12}+\frac{3}{16}=\frac{20}{48}+\frac{9}{48}=\frac{29}{48}\)

b: \(\frac{4}{15}-\frac29=\frac{12}{45}-\frac{10}{45}=\frac{12-10}{45}=\frac{2}{45}\)

a. 50-17+2-50+15

= (50-50)+(15+2-17)

= 0+0 = 0

a.0

b.79
c.16400
d.10

Haha

a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)

= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)

= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)

= 1 - 1 + \(\dfrac{2022}{2023}\)

= \(\dfrac{2022}{2023}\) 

b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)

= \(\dfrac{33}{11}\)

= 3 

c, 2000 + { 20 - [ 4.2022⁰ - (3² + 5):2] }

= 2000 + { 20 - [ 4.1 - (9+5):2]}

= 2000 + { 20 - [ 4 - 14 : 2 ]}

= 2000 + { 20 - [ 4 -7]}

= 2000 + { 20 - (-3)}

= 2000 + 23

= 2023

a: \(\frac{7}{22}+\frac{-15}{23}+\frac{2022}{2023}+\frac{-8}{23}+\frac{15}{22}\)

\(=\left(\frac{7}{22}+\frac{15}{22}\right)+\left(-\frac{15}{23}-\frac{8}{23}\right)+\frac{2022}{2023}\)

\(=1-1+\frac{2022}{2023}=\frac{2022}{2023}\)

b: \(-\frac{2}{11}+5\frac56\left(14\frac15-11\frac15\right):5\frac12\)

\(=-\frac{2}{11}+\frac{35}{6}\left(14+\frac15-11-\frac15\right):\frac{11}{2}\)

\(=\frac{-2}{11}+\frac{35}{6}\cdot\frac{2}{11}\cdot3=\frac{-2}{11}+\frac{35}{11}=\frac{33}{11}=3\)

c: \(2000+\left\lbrace20-\left\lbrack4\cdot2022^0-\left(3^2+5\right):2\right\rbrack\right\rbrace\)

\(=2000+20-\left\lbrack4\cdot1-14:2\right\rbrack\)

=2020-(4-7)

=2020-(-3)

=2023

a) \(3.5^2+15.2^2-26\div2\)

= 3.25 + 15.4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) \(5^3.2-100\div4+2^3.5\)

= 125.2 - 25 + 8.5

= 250 - 25 + 40

= 225 + 40

= 265

c)\(6^2\div9+50.2-3^3.33\)

= 36 : 9 + 100 - 9.33

= 4 + 100 - 297

= 104 - 297

= -193

d)\(3^2.5+2^3.10-81\div3\)

= 9.5 + 8.10 - 27

= 45 + 80 - 27

= 125 - 27

= 98

e) \(5^{13}\div5^{10}-25.2^2\)

= 5³ - 25.4

= 125 - 100

= 25

f) \(20\div2^2+5^9\div5^8\)

= 20 : 4 + 5

= 5 + 5

= 10

d: \(\dfrac{1}{27}:\left(-\dfrac{1}{3}\right)^2+75\%\cdot\left(-\dfrac{2^2}{3}\right)\)

\(=\dfrac{1}{27}:\dfrac{1}{9}+\dfrac{3}{4}\cdot\dfrac{-4}{3}\)

\(=\dfrac{1}{3}-1\)

\(=-\dfrac{2}{3}\)