a.\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
=\(\left(\frac{6}{14}+\frac{7}{14}\right)^2\)
=\(\left(\frac{13}{14}\right)^2\)
=\(\frac{13^2}{14^2}\)
=\(\frac{169}{196}\)
b.\(\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
=\(\left(\frac{9}{12}-\frac{10}{12}\right)^2\)
=\(\left(\frac{-1}{12}\right)^2\)
=\(\frac{-1^2}{12^2}\)
=\(\frac{1}{144}\).
c.Phần C bn viết lại đề bài đi,mk ko hiểu
d.\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)
=\(\frac{-10^5}{3^5}.\left(\frac{-6^4}{5^4}\right)\)
=\(\frac{-100000}{243}.\frac{1296}{625}\)
=\(\frac{-2560}{3}\)
Không biết đúng ko nữa
c.\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{25^2.\left(5.4\right)^4}{25^5.4^5}\)
=\(\frac{1.\left(5\right)^4.\left(4\right)^4}{25^3.4^5}\)
=\(\frac{25^2.1}{25^3.4}\)
=\(\frac{1}{25.4}\)
=\(\frac{1}{100}\).
Đề bài phần c là thế này ak Nguyễn Bình Minh:
\(\frac{5^4.20^4}{25^5.4^5}\) ak??
