b) $\frac{5}{9} \cdot \frac{7}{15}+\frac{5}{9} \cdot \frac{8}{15}-\frac{1}{3}$

Q: b) $\frac{5}{9} \cdot \frac{7}{15}+\frac{5}{9} \cdot \frac{8}{15}-\frac{1}{3}$

<p><span class="math-tex">$\begin{array}{l}\frac{5}{9} \cdot \frac{7}{15}+\frac{5}{9} \cdot \frac{8}{15}-\frac{1}{3} \\ =\frac{5}{9} \cdot\left(\frac{7}{15}+\frac{8}{15}\right)-\frac{1}{3} \\ =\frac{5}{9} \cdot 1-\frac{1}{3} \\ =\frac{5}{9}-\frac{1}{3} \\ =\frac{5}{9}-\frac{3}{9} \\ =\frac{2}{9}\end{array}$</span></p>

Question

Accepted Answer

$\begin{array}{l}\frac{5}{9} \cdot \frac{7}{15}+\frac{5}{9} \cdot \frac{8}{15}-\frac{1}{3} \ =\frac{5}{9} \cdot\left(\frac{7}{15}+\frac{8}{15}\right)-\frac{1}{3} \ =\frac{5}{9} \cdot 1-\frac{1}{3} \ =\frac{5}{9}-\frac{1}{3} \ =\frac{5}{9}-\frac{3}{9} \ =\frac{2}{9}\end{array}$

Đề thi cuối kì 2 (CT) 23-24 - Q. Bắc Từ Liêm - TP. Hà Nội - MĐ 4918