Giải bài 1 trang 169 - SGK Toán lớp 4

a)$(\dfrac{6}{11}+ \dfrac{5}{11}) \times \dfrac{3}{7};$       

Cách 1: 

$(\dfrac{6}{11}+ \dfrac{5}{11}) \times \dfrac{3}{7}\\ =1\times \dfrac{3}{7}=\dfrac{3}{7}$

Cách 2:

$(\dfrac{6}{11}+ \dfrac{5}{11}) \times \dfrac{3}{7}\\ =\dfrac{6}{11} \times \dfrac{3}{7}+ \dfrac{5}{11} \times \dfrac{3}{7}\\=\dfrac{18}{77}+\dfrac{15}{77}=\dfrac{33}{77}=\dfrac{3}{7}$

b) $\dfrac{3}{5} \times \dfrac{7}{9} - \dfrac{3}{5} \times \dfrac{2}{9}$

Cách 1: 

$\dfrac{3}{5} \times \dfrac{7}{9} - \dfrac{3}{5} \times \dfrac{2}{9}\\=​​\dfrac{21}{45}-\dfrac{6}{45}=\dfrac{15}{45}= \dfrac{1}{3}$

Cách 2: 

$\dfrac{3}{5} \times \dfrac{7}{9} - \dfrac{3}{5} \times \dfrac{2}{9}\\=​​\dfrac{3}{5} \times (\dfrac{7}{9}-\dfrac{2}{9})\\=\dfrac{3}{5} \times \dfrac{5}{9}= \dfrac{15}{45}= \dfrac{1}{3}$

c) $(\dfrac{6}{7}-\dfrac{4}{7}) : \dfrac{2}{5};$     

Cách 1:

   $(\dfrac{6}{7}-\dfrac{4}{7}) : \dfrac{2}{5}\\=\dfrac{2}{7} : \dfrac{2}{5}\\=\dfrac{2}{7} \times\dfrac{5}{2}=\dfrac{10}{14}=\dfrac{5}{7}$

Cách 2:

$(\dfrac{6}{7}-\dfrac{4}{7}) : \dfrac{2}{5}\\=\dfrac{6}{7} : \dfrac{2}{5}-\dfrac{4}{7} : \dfrac{2}{5}\\=\dfrac{6}{7} \times\dfrac{5}{2}-\dfrac{4}{7} \times\dfrac{5}{2}\\=\dfrac{30}{14}-\dfrac{20}{14}=\dfrac{10}{14}=\dfrac{5}{7}$

d) $\dfrac{8}{15} : \dfrac{2}{11} + \dfrac{7}{15} : \dfrac{2}{11}$

Cách 1:

$\dfrac{8}{15} : \dfrac{2}{11} + \dfrac{7}{15} : \dfrac{2}{11} \\=\dfrac{8}{15} \times\frac{11}{2} + \dfrac{7}{15} \times\dfrac{11}{2} \\=\dfrac{88}{30} +\dfrac{77}{30} =\dfrac{165}{30} =\dfrac{11}{2}$

Cách 2:

$\dfrac{8}{15} : \dfrac{2}{11} + \dfrac{7}{15} : \dfrac{2}{11} \\=(\dfrac{8}{15} + \frac{7}{15}) : \dfrac{2}{11}\\=1 \times\dfrac{11}{2} \\=\dfrac{11}{2}$