Question

[Lê Đặng Quỳnh Như]

Hãy viết phần số \(\frac{31}{32}\)thành tổng các phân số có tử số là 1 nhưng có mẫu số khác nhau .

Answer 1

[thien ty tfboys]

\(\frac{31}{32}=\frac{1}{32}+\frac{30}{32}=\frac{1}{32}+\frac{15}{16}=\frac{1}{32}+\frac{1}{16}+\frac{14}{16}\)

\(\Rightarrow\frac{31}{32}=\frac{1}{32}+\frac{1}{16}+\frac{7}{8}=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{6}{8}\)

\(\Rightarrow\frac{31}{32}=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{3}{4}=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{2}{4}\)

\(=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}\)

​Vậy :\(\frac{31}{32}=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}\)

Answer 2

[thien ty tfboys]

\(\frac{31}{32}=\frac{16+8+4+2+1}{32}=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)