\(\frac{10n+7}{5n-1}=\frac{2\left(5n-1\right)+9}{5n-1}=2+\frac{9}{5n-1}\)

Vậy \(5n-1\inƯ_{\left(9\right)}=\left\{\pm1;\pm3;\pm9\right\}\)

Bạn tự thay vào tìm n nhé^^

học tốt

Ta có: \(A=\frac{10n+7}{5n-1}=\frac{\left(10n-2\right)+9}{5n-1}=\frac{2\cdot\left(5n-1\right)+9}{5n-1}=2+\frac{9}{5n-1}\)

Để A nguyên => \(\frac{9}{5n-1}\inℤ\Rightarrow\left(5n-1\right)\inƯ\left(9\right)\)

\(\Leftrightarrow\left(5n-1\right)\in\left\{\pm1;\pm3;\pm9\right\}\)

\(\Leftrightarrow5n\in\left\{-8;-2;0;2;4;10\right\}\)

\(\Leftrightarrow n\in\left\{-\frac{8}{5};-\frac{2}{5};0;\frac{2}{5};\frac{4}{5};2\right\}\)

Mà n nguyên

=> \(n\in\left\{0;2\right\}\)

Ta có : \(2bd=c.\left(b+d\right)\)

\(\left(a+c\right).d=c.\left(b+d\right)\)

\(ad+cd=bc+cd\)

\(ad=bc\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

Thay \(2b=a+c\) vào ta được: 

\(2bd=c\left(b+d\right)\)

\(\Leftrightarrow\left(a+c\right)d=c\left(b+d\right)\)

\(\Leftrightarrow ad+cd=bc+cd\)

\(\Leftrightarrow ad=bc\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)

tất cả trên \(32^{16}\)hay mỗi \(\frac{16^4}{32^{16}}\)???

bạn tách ra cơ số nguyên tố 

Đặt thừa số chung 

Rút gọn tử với mẫu

           Bài làm :

Ta có :

\(b=\left(-1\right)^{2020}.\left(\frac{2}{5}\right)^3.\left(\frac{15}{4}\right)^2\div\left(\frac{15^2}{2^4}\right).\left(\frac{2}{5}\right)^3\)

\(b=1.\left[\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^3\right].\left[\left(\frac{15^2}{4^2}\right)\div\frac{15^2}{2^4}\right]\)

\(b=\left(\frac{2}{5}\right)^6.\left(\frac{15^2}{16}\div\frac{15^2}{16}\right)\)

\(b=\left(\frac{2}{5}\right)^6.1=\left(\frac{2}{5}\right)^6\)