Question

[Lương Nguyễn Gia Huy]

 

Answer 1

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

`#3107.101107`

`a)`

\(\dfrac{23}{7}+\dfrac{4}{3}-\dfrac{9}{7}+\dfrac{10}{6}\\ =\left(\dfrac{23}{7}-\dfrac{9}{7}\right)+\left(\dfrac{4}{3}+\dfrac{10}{6}\right)\\ =\dfrac{14}{7}+\left(\dfrac{4}{3}+\dfrac{5}{3}\right)\\ =\dfrac{14}{7}+\dfrac{9}{3}\\ =2+2\\ =4\)

`b)`

\(\left(\dfrac{5}{8}-\dfrac{\sqrt{9}}{12}\right)\div\dfrac{3}{4}+\dfrac{11}{8}\div\dfrac{3}{4}\)

\(=\left(\dfrac{5}{8}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{8}\cdot\dfrac{3}{4}\)

\(=\left(\dfrac{5}{8}-\dfrac{1}{4}\right)\cdot\dfrac{4}{3}+\dfrac{11}{8}\cdot\dfrac{4}{3}\)

\(=\dfrac{4}{3}\cdot\left(\dfrac{5}{8}-\dfrac{1}{4}+\dfrac{11}{8}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{16}{8}-\dfrac{1}{4}\right)\\ =\dfrac{4}{3}\cdot\left(2-\dfrac{1}{4}\right)\\ =\dfrac{4}{3}\cdot\dfrac{7}{4}\\ =\dfrac{7}{3}\)

`c)`

\(\left(0,\left(3\right)+\dfrac{\left|-2\right|}{3}\right)\div\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\\ =\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\div\dfrac{5}{4}-\left(8+9\right)^0\\ =\dfrac{3}{3}\div\dfrac{5}{4}-17^0\\ =1\div\dfrac{5}{4}-1\\ =\dfrac{5}{4}-1\\ =\dfrac{1}{4}\)

Answer 2

[nguyễn xuân tùng]

đù lớp 1 dữ