Tìm tất cả các số nguyên n để A = 10n+ 7/ 5n-1
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\(\frac{10n+7}{5n-1}=\frac{2\left(5n-1\right)+9}{5n-1}=2+\frac{9}{5n-1}\)
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
=> \(n\in\left\{0;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(2bd=c\left(b+d\right)\)
\(\Leftrightarrow\left(a+c\right)d=c\left(b+d\right)\)
\(\Leftrightarrow ad+cd=bc+cd\)
\(\Leftrightarrow ad=bc\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tất cả trên \(32^{16}\)hay mỗi \(\frac{16^4}{32^{16}}\)???
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm :
Ta có :
\(b=\left(-1\right)^{2020}.\left(\frac{2}{5}\right)^3.\left(\frac{15}{4}\right)^2\div\left(\frac{15^2}{2^4}\right).\left(\frac{2}{5}\right)^3\)
\(b=1.\left[\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^3\right].\left[\left(\frac{15^2}{4^2}\right)\div\frac{15^2}{2^4}\right]\)
\(b=\left(\frac{2}{5}\right)^6.\left(\frac{15^2}{16}\div\frac{15^2}{16}\right)\)
\(b=\left(\frac{2}{5}\right)^6.1=\left(\frac{2}{5}\right)^6\)