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\(6=\frac{6}{1}=\frac{12}{2}=\frac{24}{4}=\frac{48}{8}=\frac{96}{16}\)
\(5:9=\frac{5}{9}=\frac{10}{18}=\frac{20}{36}=\frac{40}{72}=\frac{80}{144}\)
\(1=\frac{3}{3}=\frac{5}{5}=\frac{7}{7}=\frac{9}{9}=\frac{11}{11}\)
\(0=\frac{0}{7}=\frac{0}{10}=\frac{0}{13}=\frac{0}{16}=\frac{0}{19}\)
Chúc em hok tốt!!!
a) \(\frac{3}{5}\times y+\frac{1}{2}:\frac{5}{3}-\frac{5}{4}=\frac{1}{2}\times\frac{1}{3}\)
\(\Rightarrow\frac{3}{5}\times y+\frac{3}{10}-\frac{5}{4}=\frac{1}{6}\)
\(\Rightarrow\frac{3}{5}\times y+\left(-\frac{19}{20}\right)=\frac{1}{6}\)
\(\Rightarrow\frac{3}{5}\times y=\frac{67}{60}\)
\(\Rightarrow y=\frac{67}{36}\)
b) \(\frac{4}{5}:y+\frac{1}{4}\times\frac{1}{6}-\frac{1}{2}=\frac{1}{3}\times\frac{5}{2}\)
\(\Rightarrow\frac{4}{5}:y+\frac{1}{24}-\frac{1}{2}=\frac{5}{6}\)
\(\Rightarrow\frac{4}{5}:y+\left(-\frac{11}{24}\right)=\frac{5}{6}\)
\(\Rightarrow\frac{4}{5}:y=\frac{5}{6}+\frac{11}{24}=\frac{31}{24}\)
\(\Rightarrow y=\frac{4}{5}:\frac{31}{24}=\frac{96}{155}\)
c) \(\frac{3}{5}\times y-\frac{4}{5}:3+\frac{1}{12}=\frac{3}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}+\frac{1}{12}=\frac{17}{10}\)
\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}=\frac{97}{60}\)
\(\Rightarrow\frac{3}{5}\times y=\frac{113}{60}\)
\(\Rightarrow y=\frac{113}{36}\)
a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
- a B,b D
- a \(\frac{1}{2}\)b \(\frac{2}{5}\)
- \(\frac{2}{3};\frac{10}{17};\frac{5}{11};\frac{4}{9}\)
- a\(\frac{5}{12}\)b\(\frac{97}{36}\)
B
lười ko tốt cho sức khỏe đou :D