Tính một cách hợp lí: a) 11/4*-5/9*8/33 b) -5/6*4/19+-7/12*4/19-40/57

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

Bài 40 trang 40 SBT Toán 6 Tập 2:

Tính một cách hợp lí:

a) $\frac{11}{4}\cdot \frac{-5}{9}\cdot \frac{8}{33}$;                     b) $\frac{-5}{6}\cdot \frac{4}{19}+\frac{-7}{12}\cdot \frac{4}{19}-\frac{40}{57}$;

c) $\left(\frac{23}{41}-\frac{15}{82}\right)\cdot \frac{41}{15}$                  d) $9\cdot \left(\frac{151515}{171717}-\frac{131313}{181818}\right)$;

e) $\frac{-13}{8}\cdot \left(\frac{8}{13}+\frac{32}{28}\right)-\frac{15}{7}$         g) $\frac{2^2}{1.3}\cdot \frac{3^2}{2.4}\cdot \frac{4^2}{3.5}\cdot \frac{5^2}{4.6}\cdot \frac{6^2}{5.7}$.

Lời giải

a) $\frac{11}{4}\cdot \frac{-5}{9}\cdot \frac{8}{33}=\frac{11\cdot (-5)\cdot 2\cdot 4}{4\cdot 9\cdot 3\cdot 11}=\frac{-10}{27}$.

b) $\frac{-5}{6}\cdot \frac{4}{19}+\frac{-7}{12}\cdot \frac{4}{19}-\frac{40}{57}=\frac{4}{19}\left(\frac{-5}{6}+\frac{-7}{12}\right)-\frac{40}{57}$ 

$=\frac{4}{19}\left(\frac{-10}{12}+\frac{-7}{12}\right)-\frac{40}{57}=\frac{4}{19}\cdot \frac{-17}{12}-\frac{40}{57}$

$=\frac{1}{19}.\frac{-17}{3}-\frac{40}{57}=\frac{-17}{57}-\frac{40}{57}=\frac{-57}{57}=-1.$

c) $\left(\frac{23}{41}-\frac{15}{82}\right)\cdot \frac{41}{15}=\frac{23}{41}\cdot \frac{41}{15}-\frac{15}{82}\cdot \frac{41}{15}$

$=\frac{23}{15}-\frac{1}{2}=\frac{46}{30}-\frac{15}{30}=\frac{31}{30}$.

d) $9\cdot \left(\frac{151515}{171717}-\frac{131313}{181818}\right)=9\cdot \left(\frac{15.10101}{17.10101}-\frac{13.10101}{18.10101}\right)$

$=9\cdot \left(\frac{15}{17}-\frac{13}{18}\right)=9.\left(\frac{270}{306}-\frac{221}{306}\right)=9\cdot \frac{49}{306}=\frac{49}{34}$.

e) $\frac{-13}{8}\cdot \left(\frac{8}{13}+\frac{32}{28}\right)-\frac{15}{7}=\frac{-13}{8}\cdot \left(\frac{8}{13}+\frac{8}{7}\right)-\frac{15}{7}$ 

$=\frac{-13}{8}\cdot \frac{8}{13}-\frac{13}{8}\cdot \frac{8}{7}-\frac{15}{7}=-1-\frac{13}{7}-\frac{15}{7}$

$=-1-\left(\frac{13}{7}+\frac{15}{7}\right)=-1-\frac{28}{7}=\frac{-7}{7}-\frac{28}{7}=\frac{-35}{7}=-5$.

g) $\frac{2^2}{1.3}\cdot \frac{3^2}{2.4}\cdot \frac{4^2}{3.5}\cdot \frac{5^2}{4.6}\cdot \frac{6^2}{5.7}$

$=\frac{\mathrm{2.2.3.3.4.4.5.5.6.6}}{\mathrm{1.3.2.4.3.5.4.6.5.7}}$

$=\frac{\mathrm{2.2.3.3.4.4.5.5.6.6}}{\mathrm{1.2.3.3.4.4.5.5.6.7}}$$=\frac{2.6}{7}=\frac{12}{7}\text{. }$