Tính: a) 3/5+3/27-3/9-3/11/4/5+4/27-4/9-4/11; b)5-5/3-5/27/8-8/3-8/27:15+15/121-15/11/16+16/121-16/11

Giải SBT Toán 6 Cánh diều Bài 4: Phép nhân, phép chia phân số  

Bài 50 trang 42 SBT Toán 6 Tập 2:

Tính:

a) $\frac{\frac{3}{5}+\frac{3}{27}-\frac{3}{9}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{27}-\frac{4}{9}-\frac{4}{11}}$;

b) $\frac{5-\frac{5}{3}-\frac{5}{27}}{8-\frac{8}{3}-\frac{8}{27}}:\frac{15+\frac{15}{121}-\frac{15}{11}}{16+\frac{16}{121}-\frac{16}{11}}$;

c) $\frac{1}{2}:\left(\frac{-3}{2}\right):\frac{4}{3}:\left(\frac{-5}{4}\right):\frac{6}{5}:\left(\frac{-7}{6}\right):\dots :\left(\frac{-101}{100}\right)$.

Lời giải

a) $\frac{\frac{3}{5}+\frac{3}{27}-\frac{3}{9}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{27}-\frac{4}{9}-\frac{4}{11}}$

$=\frac{3\cdot \left(\frac{1}{5}+\frac{1}{27}-\frac{1}{9}-\frac{1}{11}\right)}{4\cdot \left(\frac{1}{5}+\frac{1}{27}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{3}{4}$.

b) $\frac{5-\frac{5}{3}-\frac{5}{27}}{8-\frac{8}{3}-\frac{8}{27}}:\frac{15+\frac{15}{121}-\frac{15}{11}}{16+\frac{16}{121}-\frac{16}{11}}$

$=\frac{5\cdot \left(1-\frac{1}{3}-\frac{1}{27}\right)}{8\cdot \left(1-\frac{1}{3}-\frac{1}{27}\right)}:\frac{15\cdot \left(1+\frac{1}{121}-\frac{1}{11}\right)}{16\cdot \left(1+\frac{1}{121}-\frac{1}{11}\right)}$

$=\frac{5}{8}:\frac{15}{16}=\frac{5}{8}\cdot \frac{16}{15}=\frac{2}{3}$.

c) $\frac{1}{2}:\left(\frac{-3}{2}\right):\frac{4}{3}:\left(\frac{-5}{4}\right):\frac{6}{5}:\left(\frac{-7}{6}\right):\dots :\left(\frac{-101}{100}\right)$

$=\frac{1}{2}\cdot \left(\frac{2}{-3}\right)\cdot \frac{3}{4}\cdot \left(\frac{4}{-5}\right)\cdot \frac{5}{6}.\left(\frac{6}{-7}\right)\cdot \cdot \cdot \cdot \cdot \left(\frac{100}{-101}\right)$

$=\frac{1}{2}\cdot \frac{-2}{3}\cdot \frac{3}{4}\cdot \frac{-4}{5}\cdot \frac{5}{6}\cdot \frac{-6}{7}\cdot \cdot \cdot \cdot \cdot \frac{-100}{101}$

$=\frac{1}{101}\text{.}$  (tích có 50 thừa số dương và 50 thừa số âm là số dương).