M=\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\frac{2x-5y}{x-3y}
=\frac{2.10y-5y}{10y-3y}=\frac{15}{7}
Câu 2
\(\frac{x}{y}\)nha bạn
a) Để giá trị của biểu thức B xác định
\(\Leftrightarrow\hept{\begin{cases}x^2-7x\ne0\\x^2+7x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-7\right)\ne0\\x\left(x+7\right)\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne\pm7\end{cases}}}\)
Vậy ................
b) \(B=\left(\frac{7x+1}{x^2-7x}+\frac{7x-1}{x^2+7x}\right):\frac{x^2+1}{x^2-49}\)
\(B=\left(\frac{7x+1}{x\left(x-7\right)}+\frac{7x-1}{x\left(x+7\right)}\right).\frac{\left(x+7\right)\left(x-7\right)}{x^2+1}\)
\(B=\frac{\left(7x+1\right)\left(x+7\right)+\left(7x-1\right)\left(x-7\right)}{x\left(x-7\right)\left(x+7\right)}.\frac{\left(x-7\right)\left(x+7\right)}{x^2+1}\)
\(B=\frac{7x^2+49x+x+7+7x^2-49x-x+7}{x\left(x^2+1\right)}\)
\(B=\frac{14x^2+14}{x\left(x^2+1\right)}\)
\(B=\frac{14}{x}\)
c) Để \(B\inℤ\)
\(\Leftrightarrow\frac{14}{x}\inℤ\)
\(\Leftrightarrow x\inƯ\left(14\right)\)
\(Ư\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)
Vậy ...................