\(\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{40.10^3+7.10^3}=\frac{33.10^3}{10^3.47}=\frac{33}{47}\)

\(\frac{3774}{5217}=\frac{34}{47}\)

Do đó VT<VP

33.10³/2³.5.10³+7000<3774/5217

\(\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(2^3.5+7\right)}=\frac{33}{8.5+7}=\frac{33}{47}\)

\(\frac{3774}{5217}=\frac{3774:111}{5217:111}=\frac{34}{47}\)

Vì \(\frac{33}{47}< \frac{34}{47}\Rightarrow\frac{33.10^3}{2^3.5.10^3+7000}< \frac{3774}{5217}\)

\(\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{8.5.10^3+7.10^3}\)

=\(\frac{33.10^3}{10^3\left(40+7\right)}=\frac{33}{47}\)

\(\frac{3774}{5217}=\frac{111.34}{111.47}=\frac{34}{47}\)

Vậy: \(\frac{3774}{5217}>\frac{33.10^3}{2^3.5.10^3+7000}\)

u3hfhfhfjjfjfjFegyvdydvvydcvyvyddvvdvyrvvyvrbrvyfvybrv bbcbfbfbchrubdvh dvubudvbudvbvudbudvbudvubevbudvbudvubdvbduburvubrvubrvubruvbubrvburvbururvubvrbyrvbu  jjfhfhvhvvhvhvhhfhffhjvjf(&h&7

Phân số 1 lớn hơn

\(\frac{33\cdot10^3}{2^3\cdot5\cdot10^3+7000}=\frac{33\cdot10^3}{40\cdot10^3+7\cdot10^3}=\frac{33\cdot10^3}{\left(40+7\right)\cdot10^3}=\frac{33}{47}\)

\(\frac{3774}{5217}=\frac{2\cdot3\cdot17\cdot37}{3\cdot37\cdot47}=\frac{2\cdot17}{47}=\frac{34}{47}\)

Vì 33 < 34 => \(\frac{33}{47}< \frac{34}{47}\)hay \(\frac{33\cdot10^3}{2^3\cdot5\cdot10^3+7000}< \frac{3774}{5217}\)

Áp dụng tính chất cơ bản của phân số để rút gọn:

\(\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(40+7\right)}=\frac{33}{47}\) (1)

\(\frac{3774}{5217}=\frac{3774\div111}{5217\div111}=\frac{34}{47}\) (2)

Từ (1) và (2) ta có: \(\frac{3774}{5217}>\frac{33.10^3}{2^3.5.10^3+7000}\)

\(\frac{33.10^3}{2^3.5.10^3+7000}\)= \(\frac{33.10^3}{2^3.5.10^3+7.10^3}\)

                                     =\(\frac{33.10^3}{10^3.\left(2^3.5+7\right)}\)

                                     =\(\frac{33.10^3}{10^3.47}\)

                                     =\(\frac{33}{47}\)