a) $\mathrm{A}=\frac{-3}{5} \cdot \frac{7}{13}-\frac{3}{5}: \frac{13}{6}+1 \frac{3}{5}$

Q: a) $\mathrm{A}=\frac{-3}{5} \cdot \frac{7}{13}-\frac{3}{5}: \frac{13}{6}+1 \frac{3}{5}$

<p><span class="math-tex">$\begin{array}{l}A=\frac{-3}{5} \cdot \frac{7}{13}-\frac{3}{5}: \frac{13}{6}+1 \frac{3}{5} \\ A=\frac{-3}{5} \cdot \frac{7}{13}+\frac{-3}{5} \cdot \frac{6}{13}+1 \frac{3}{5} \\ A=\frac{-3}{5} \cdot\left(\frac{7}{13}+\frac{6}{13}\right)+1 \frac{3}{5} \\ A=\frac{-3}{5} \cdot \frac{13}{13}+1 \frac{3}{5} \\ A=\frac{-3}{5}+1+\frac{3}{5} \\ A=\left(\frac{-3}{5}+\frac{3}{5}\right)+1 \\ A=1 \\ V a ̣ y \\ A=1\end{array}$</span></p>

Question

Accepted Answer

$\begin{array}{l}A=\frac{-3}{5} \cdot \frac{7}{13}-\frac{3}{5}: \frac{13}{6}+1 \frac{3}{5} \ A=\frac{-3}{5} \cdot \frac{7}{13}+\frac{-3}{5} \cdot \frac{6}{13}+1 \frac{3}{5} \ A=\frac{-3}{5} \cdot\left(\frac{7}{13}+\frac{6}{13}\right)+1 \frac{3}{5} \ A=\frac{-3}{5} \cdot \frac{13}{13}+1 \frac{3}{5} \ A=\frac{-3}{5}+1+\frac{3}{5} \ A=\left(\frac{-3}{5}+\frac{3}{5}\right)+1 \ A=1 \ V a ̣ y \ A=1\end{array}$

Đề thi HSG (CT) 18-19 - H.Thái Thụy - Thái Bình - MĐ 6191