Question

Bài 40. Tính:                                                                                                                                                Bài 41. Tính:                                                                                                                    

Answer 1

Bài 40 :

\(a,\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{13^2}{14^2}=\dfrac{169}{196}\)

\(b,\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{-1}{12}\right)^2=\dfrac{\left(-1\right)^2}{12^2}=\dfrac{1}{144}\)

\(c,\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.\left(5.4\right)^4}{\left(5^2\right)^5.4^5}=\dfrac{5^4.5^4.4^4}{5^{10}.4^5}=\dfrac{5^8.4^4}{5^{10}.4^5}=\dfrac{1}{5^2.4}=\dfrac{1}{100}\)

\(d,\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-10}{3}\right).\left(\dfrac{-6}{5}\right)^4=\left(\dfrac{-10}{3}.\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)=4^4.\dfrac{-10}{3}=256.\dfrac{-10}{3}=\dfrac{-2560}{3}\)

 

Answer 2

Bài 41 :

\(a,\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right).\left(\dfrac{16}{20}-\dfrac{15}{20}\right)^2=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)

\(b,2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3=2:\left(\dfrac{-1}{6}\right)^3=2:\dfrac{-1}{216}=-432\)