Câu hỏi của Edogawa Conan

Question

Tính hợp lýa) $\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2$b) $\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)$

Phạm Thị Thùy Linh · Accepted Answer

$a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4$$=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}$$b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)$$=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)$$=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}$$=\frac{100.12}{123.16}+\frac{23.15}{123.20}$$=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}$$=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}$$=\frac{123}{164}=\frac{3}{4}$Câu trả lời: $a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4$$=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}$$b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)$$=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)$$=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}$$=\frac{100.12}{123.16}+\frac{23.15}{123.20}$$=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}$$=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}$$=\frac{123}{164}=\frac{3}{4}$

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