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a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
:V Làm sai hết rồi sai ngay từ bước đầu tiên.
\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)
\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{3}{20}\)
\(=\frac{-11}{12}\)
\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)
= \(-\frac{7}{30}\)
\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3 \left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)
=3[1/5+1/7-1/11]
----------------------- =3/4
4[1/5+1/7-1/11]
