Question

lm bài 4 thôi nha

Answer 1

Bài 4:

1) 

\(\left(\dfrac{1}{2}\right)^2-\left[1\dfrac{2}{5}\cdot\dfrac{1}{3}+\left(2,\left(6\right)-2\right)\cdot\dfrac{7}{5}\right]\\ =\dfrac{1}{4}-\left[\dfrac{7}{5}\cdot\dfrac{1}{3}+\left(\dfrac{8}{3}-2\right)\cdot\dfrac{7}{5}\right]\\ =\dfrac{1}{4}-\dfrac{7}{5}\cdot\left(\dfrac{1}{3}+\dfrac{8}{3}-2\right)\\ =\dfrac{1}{4}-\dfrac{7}{5}\\ =-\dfrac{23}{20}\)

2) 

\(\left(4-\dfrac{19}{5}\right)^2+\dfrac{-3}{10}\cdot0,\left(6\right)^2-\dfrac{2^3}{75}\\ =\left(-\dfrac{1}{5}\right)^2+\dfrac{-3}{10}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{8}{75}\\ =\dfrac{1}{25}+\dfrac{-3}{10}\cdot\dfrac{4}{9}-\dfrac{8}{75}\\ =\dfrac{1}{25}+\dfrac{-2}{15}-\dfrac{8}{75}\\ =\dfrac{3}{75}+\dfrac{-10}{75}-\dfrac{8}{75}\\ =\dfrac{-15}{75}\\ =-\dfrac{1}{5}\) 

3) 

\(0,\left(6\right)^2\cdot9-\left(-\dfrac{1}{4}\right)^3:\dfrac{1}{64}\\ =\left(\dfrac{2}{3}\right)^2\cdot9-\left(-\dfrac{1}{64}\right):\dfrac{1}{64}\\ =\dfrac{4}{9}\cdot9+\dfrac{1}{64}:\dfrac{1}{64}\\ =4+1\\ =5\) 

4) 

\(\dfrac{3}{5}:\left(-\dfrac{1}{15}-0,1\left(6\right)\right)+\dfrac{3}{5}:\left(\dfrac{-1^6}{3}-1\dfrac{1}{15}\right)\\ =\dfrac{3}{5}:\left(-\dfrac{1}{15}-\dfrac{1}{6}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\\ =\dfrac{3}{5}:-\dfrac{7}{30}+\dfrac{3}{5}:-\dfrac{7}{5}\\ =\dfrac{3}{5}\cdot\dfrac{-30}{7}+\dfrac{3}{5}\cdot\dfrac{-5}{7}\\ =\dfrac{3}{5}\cdot\left(-\dfrac{30}{7}+\dfrac{-5}{7}\right)\\ =\dfrac{3}{5}\cdot-5=-3\)