Question

Thực hiện phép tính theo cách hợp lí :

a, \([6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1]:\left(-\frac{1}{3}-1\right)\)

b, \(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^3}\)

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Answer 1

a)\(\left[6.\left(-\frac{1}{3}\right)^2-3\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)

\(=\frac{\left[6\left(-\frac{1}{3}\right)^2+3\left(-\frac{1}{3}\right)+1\right]}{-\frac{1}{3}}-\frac{\left[6\left(-\frac{1}{3}\right)^2-3\left(-\frac{1}{3}\right)+1\right]}{-1}\)

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

\(=6\left(-\frac{1}{3}\right)+3-3+\frac{6.1}{9}+\frac{3}{3}+1\)

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

Answer 2

Answer 3

Làm ơn thì làm ơn cho trót đi ! Giúp mk nốt câu b nha !