1. tính
a, (\(\dfrac{4}{5}+\dfrac{2}{3}\\ \) ) x \(\dfrac{1}{6}\) b, ( \(\dfrac{7}{8}-\dfrac{4}{16}\\ \) ) x \(\dfrac{2}{3}\)
c, \(\dfrac{5}{9}x\dfrac{2}{4}+\dfrac{4}{6}x\dfrac{2}{4}\\ \) d, \(\dfrac{15}{3}x\dfrac{2}{6}-\dfrac{11}{7}x\dfrac{2}{6}\\ \)
2. Tính bằng hai cách
a, \(\dfrac{7}{5}+\dfrac{9}{7}+\dfrac{6}{5}+\dfrac{11}{7}\\ \) b, \(\dfrac{8}{11}+\dfrac{2}{11}+\dfrac{12}{12}+\dfrac{31}{11}\\\)
Uyên ương hữu tình, giúp đỡ một tay để mình không trôi dạt với câu hỏi khó nhằn này được không?
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Đỗ Đăng Đạt
Bài 2:
a: C1: \(\dfrac{7}{5}+\dfrac{9}{7}+\dfrac{6}{5}+\dfrac{11}{7}\)
\(=\left(\dfrac{7}{5}+\dfrac{6}{5}\right)+\left(\dfrac{11}{7}+\dfrac{9}{7}\right)\)
\(=\dfrac{13}{5}+\dfrac{20}{7}=\dfrac{91+100}{35}=\dfrac{191}{35}\)
C2: \(\dfrac{7}{5}+\dfrac{9}{7}+\dfrac{6}{5}+\dfrac{11}{7}\)
\(=\dfrac{49}{35}+\dfrac{45}{35}+\dfrac{42}{35}+\dfrac{55}{35}\)
\(=\dfrac{49+45+42+55}{35}=\dfrac{191}{35}\)
b: C1: \(\dfrac{8}{11}+\dfrac{2}{11}+\dfrac{12}{12}+\dfrac{31}{11}\)
\(=\left(\dfrac{8}{11}+\dfrac{2}{11}+\dfrac{31}{11}\right)+1\)
\(=\dfrac{41}{11}+1=\dfrac{52}{11}\)
C2: \(\dfrac{8}{11}+\dfrac{2}{11}+\dfrac{12}{12}+\dfrac{31}{11}\)
\(=\dfrac{96}{132}+\dfrac{24}{132}+\dfrac{132}{132}+\dfrac{372}{132}\)
\(=\dfrac{96+24+132+372}{132}=\dfrac{624}{132}=\dfrac{52}{11}\)
Bài 1:
a: \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\times\dfrac{1}{6}=\dfrac{12+10}{15}\times\dfrac{1}{6}\)
\(=\dfrac{22}{90}=\dfrac{11}{45}\)
b: \(\left(\dfrac{7}{8}-\dfrac{4}{16}\right)\times\dfrac{2}{3}=\left(\dfrac{14}{16}-\dfrac{4}{16}\right)\times\dfrac{2}{3}\)
\(=\dfrac{10}{16}\times\dfrac{2}{3}=\dfrac{5}{8}\times\dfrac{2}{3}=\dfrac{5}{12}\)
c: \(\dfrac{5}{9}\times\dfrac{2}{4}+\dfrac{4}{6}\times\dfrac{2}{4}\)
\(=\dfrac{2}{4}\left(\dfrac{5}{9}+\dfrac{2}{3}\right)\)
\(=\dfrac{1}{2}\times\dfrac{5+6}{9}=\dfrac{1}{2}\times\dfrac{11}{9}=\dfrac{11}{18}\)
d: \(\dfrac{15}{3}\times\dfrac{2}{6}-\dfrac{11}{7}\times\dfrac{2}{6}\)
\(=\dfrac{2}{6}\times\left(\dfrac{15}{3}-\dfrac{11}{7}\right)\)
\(=\dfrac{1}{3}\times\left(5-\dfrac{11}{7}\right)=\dfrac{1}{3}\times\dfrac{24}{7}=\dfrac{8}{7}\)