Ta có: \(\frac{2^{12} \cdot 3^{5} - 4^{6} \cdot 9^{2}}{\left(\left(\right. 2^{2} \cdot 3 \left.\right)\right)^{6} + 8^{4} \cdot 3^{5}} - \frac{5^{10} \cdot 7^{3} - 2 5^{5} \cdot 4 9^{2}}{\left(\left(\right. 125 \cdot 7 \left.\right)\right)^{3} + 5^{9} \cdot 1 4^{3}}\)

\(= \frac{2^{12} \cdot 3^{5} - 2^{12} \cdot 3^{4}}{2^{12} \cdot 3^{6} + 2^{12} \cdot 3^{5}} - \frac{5^{10} \cdot 7^{3} - 5^{10} \cdot 7^{4}}{5^{9} \cdot 7^{3} + 5^{9} \cdot 7^{3} \cdot 2^{3}}\)

\(= \frac{2^{12} \cdot 3^{4} \left(\right. 3 - 1 \left.\right)}{2^{12} \cdot 3^{5} \left(\right. 3 + 1 \left.\right)} + \frac{5^{10} \cdot 7^{3} \left(\right. 7 - 1 \left.\right)}{5^{9} \cdot 7^{3} \left(\right. 1 + 2^{3} \left.\right)} = \frac{1}{3} \cdot \frac{2}{4} + \frac{5 \cdot 6}{9} = \frac{1}{6} + \frac{10}{3} = \frac{21}{6} = \frac{7}{2}\)

1000000000000−99= \(999999999901.\)

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3

(\(\frac{9}{2.3}\) + \(\frac{9}{3.4}\) + ... + \(\frac{9}{199.200}\)).\(x\) = \(\frac{2}{5}\)

9.(\(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ... + \(\frac{1}{199.200}\)).\(x\) = \(\frac{2}{5}\)

9.(\(\frac{1}{2} - \frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{199}\) - \(\frac{1}{200}\)).\(x\) = \(\frac{2}{5}\)

9.(\(\frac{1}{2}\) - \(\frac{1}{200}\)).\(x\) = \(\frac{2}{5}\)

9.\(\frac{99}{200}\).\(x\) = \(\frac{2}{5}\)

\(\frac{891}{200}\).\(x\) = \(\frac{2}{5}\)

\(x\) = \(\frac{2}{5}\) : \(\frac{891}{200}\)

\(x\) = \(\frac{80}{891}\)

Vậy \(x = \frac{80}{891}\)

khoa học tự nhiên nhé

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