Lời giải Câu b Bài tập cuối Chương 6 (trang 27) – SGK Toán 6 Kết nối tri thức. Gợi ý: Nhóm \(\dfrac{5}{3}. \dfrac{7}{{25}} – \dfrac{5}{3}. \dfrac{7}{{25}}\) rồi tính.
Câu hỏi/Đề bài:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} – \dfrac{5}{3}.\dfrac{7}{{25}}\)
Hướng dẫn:
Nhóm \(\dfrac{5}{3}.\dfrac{7}{{25}} – \dfrac{5}{3}.\dfrac{7}{{25}}\) rồi tính.
Lời giải:
Cách 1:
\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} – \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} – \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)
Cách 2:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} – \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B = \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)