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a: \(\dfrac{2}{3}+\dfrac{7}{3}=\dfrac{9}{3}\)
\(\dfrac{7}{3}+\dfrac{2}{3}=\dfrac{9}{3}\)
=>\(\dfrac{2}{3}+\dfrac{7}{3}=\dfrac{7}{3}+\dfrac{2}{3}\)
\(\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}\)
=>\(\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{4}{5}+\dfrac{3}{5}\)
b: \(\dfrac{7}{9}+\dfrac{16}{9}=\dfrac{7+16}{9}=\dfrac{23}{9}\)
\(\dfrac{16}{9}+\dfrac{7}{9}=\dfrac{16+7}{9}=\dfrac{23}{9}\)
Do đó: \(\dfrac{7}{9}+\dfrac{16}{9}=\dfrac{16}{9}+\dfrac{7}{9}\)
a) $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ ; $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$
Vậy $\frac{1}{2} \times \frac{1}{3} = \frac{1}{3} \times \frac{1}{2}$
$\frac{3}{5} \times \frac{1}{6} = \frac{3}{{30}} = \frac{1}{{10}}$ ; $\frac{1}{6} \times \frac{3}{5} = \frac{3}{{30}} = \frac{1}{{10}}$
Vậy $\frac{3}{5} \times \frac{1}{6} = \frac{1}{6} \times \frac{3}{5}$
b) Học sinh tự thực hiện
\(\dfrac{1}{3}< \dfrac{3}{5}< \dfrac{7}{9}< \dfrac{9}{11}< \dfrac{11}{13}\)
a) \(\dfrac{3}{4}=\dfrac{3\times4}{4\times4}=\dfrac{12}{16}\)
b) \(\dfrac{1}{3}=\dfrac{1\times3}{3\times3}=\dfrac{3}{9}\)
c) \(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18}\)
a)
\(\dfrac{5}{9}< \dfrac{9}{9}\)
\(\dfrac{8}{7}>\dfrac{7}{7}\)
\(\dfrac{9}{9}=1\)
\(\dfrac{18}{4}>\dfrac{3}{4}\)
b)
\(\dfrac{2}{5},\dfrac{3}{5},\dfrac{8}{5}\)
\(\dfrac{5}{2}=\dfrac{15}{6},\dfrac{1}{6},1=\dfrac{6}{6}\rightarrow\dfrac{1}{6},\dfrac{6}{6},\dfrac{15}{6}\)
Biểu thức thứ nhất
\(\dfrac{8}{5}:\dfrac{6}{5}=\dfrac{8}{6}=\dfrac{4}{3}\)
Biểu thức thứ hai:
\(\dfrac{5}{3}-\dfrac{5}{3}=0\)
Vậy biểu thức thứ nhất lớn hơn biểu thức thứ 2
a) HS tự thực hiện
b) $\frac{5}{6}$ < 1 ; $\frac{3}{2} > 1$
$\frac{9}{{19}}$ < 1 ; $\frac{7}{7}$ = 1
$\frac{{49}}{{46}}$ > 1 ; $\frac{{32}}{{71}}$ < 1
c) Ba phân số bé hơn 1 là: $\frac{2}{7};\,\,\,\frac{{11}}{{25}};\,\,\,\frac{{37}}{{59}}$
Ba phân số lớn hơn 1 là: $\frac{7}{2};\,\,\,\frac{{15}}{7};\,\,\,\,\frac{{33}}{{12}}$
Ba phân số bằng 1 là: $\frac{9}{9};\,\,\,\,\frac{{25}}{{25}};\,\,\,\,\frac{{47}}{{47}}$
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?