Số các số nguyên x để \[\frac{{5x}}{3}:\frac{{10{x^2} + 5x}}{{21}}\] có giá trị là số nguyên là

\[\frac{{28}}{{15}}.\frac{1}{{{4^2}}}.3 + \left( {\frac{8}{{15}} - \frac{{69}}{{60}}.\frac{5}{{23}}} \right):\frac{{51}}{{54}}\]

\( = \frac{{28.1.3}}{{{{15.4}^2}}} + \left( {\frac{8}{{15}} - \frac{{23.3}}{{4.3.5}}.\frac{5}{{23}}} \right).\frac{{54}}{{51}}\)

\(\begin{array}{l} = \frac{{7.4.1.3}}{{3.5.4.4}} + \left( {\frac{8}{{15}} - \frac{1}{4}} \right).\frac{{54}}{{51}}\\ = \frac{7}{{20}} + \left( {\frac{{32}}{{60}} - \frac{{15}}{{60}}} \right).\frac{{54}}{{51}}\\ = \frac{7}{{20}} + \frac{{17}}{{60}}.\frac{{54}}{{51}}\\ = \frac{7}{{20}} + \frac{{17}}{{6.10}}.\frac{{6.3.3}}{{17.3}}\\ = \frac{7}{{20}} + \frac{3}{{10}}\\ = \frac{7}{{20}} + \frac{6}{{20}}\\ = \frac{{13}}{{20}}\end{array}\)

Đáp án cần chọn là: D

\[B = \frac{{{2^2}}}{3} \cdot \frac{{{3^2}}}{8} \cdot \frac{{{4^2}}}{{15}} \cdot \frac{{{5^2}}}{{24}} \cdot \frac{{{6^2}}}{{35}} \cdot \frac{{{7^2}}}{{48}} \cdot \frac{{{8^2}}}{{63}} \cdot \frac{{{9^2}}}{{80}}\]

\[ = \frac{{2.2}}{{1.3}} \cdot \frac{{3.3}}{{2.4}} \cdot \frac{{4.4}}{{3.5}} \cdot \frac{{5.5}}{{4.6}} \cdot \frac{{6.6}}{{5.7}} \cdot \frac{{7.7}}{{6.8}} \cdot \frac{{8.8}}{{7.9}} \cdot \frac{{9.9}}{{8.10}}\]

\[ = \frac{{2.3.4.5.6.7.8.9}}{{1.2.3.4.5.6.7.8.}} \cdot \frac{{2.3.4.5.6.7.8.9}}{{3.4.5.6.7.8.9.10}}\]

\[ = \frac{9}{1} \cdot \frac{2}{{10}} = \frac{{9.2}}{{1.10}} = \frac{9}{5}\]

Đáp án cần chọn là: A