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a) \(\frac{9}{33-3}=\frac{1}{3}\)
b) \(\frac{7}{100+6\times100}=\frac{1}{100}\)
c) \(\frac{11\times22+33\times36+55\times60}{22\times24+66\times72+110\times120}=\frac{1}{4}\)
d) \(\frac{9^4\times27^5\times3^6\times4^4}{3^8\times81^4\times243\times8^2}=4\)
e) \(\frac{199919991999}{200020002000}=\frac{1999}{2000}\)
\(H=\frac{4116-14}{10290-35}=\frac{14.294-14}{35.294-35}=\frac{14.\left(294-1\right)}{35.\left(294-1\right)}=\frac{14.293}{35.293}=\frac{2}{5}\)
\(K=\frac{29.101-101}{2.19.101+4.101}=\frac{101.\left(29-1\right)}{101.\left(38+4\right)}=\frac{28}{42}=\frac{2}{3}\)
\(I=\frac{1313-1717}{303}=\frac{13.101-17.101}{3.101}=\frac{101.\left(13-17\right)}{3.101}=\frac{-4}{3}\)
\(M=\frac{12-24.3}{1-35}=\frac{12-12.2.3}{-34}=\frac{12.\left(1-6\right)}{-34}=\frac{-60}{-34}=\frac{30}{17}\)
\(H\)\(=\) \(\frac{4116-14}{10290-35}\)
\(=\) \(\frac{4102}{10255}\)
\(=\) \(\frac{4102:2051}{10255:2051}\)
\(=\) \(\frac{2}{5}\)
\(K=\frac{2929-101}{2.1919+404}\)
\(=\) \(\frac{2828}{4242}\)
\(=\) \(\frac{2828:1414}{4242:1414}\)
\(=\) \(\frac{2}{3}\)
\(M=\frac{12-24.3}{1-35}\)
\(=\) \(\frac{-60}{-34}\)
\(=\) \(\frac{60}{34}\)
\(=\) \(\frac{30}{17}\)
:D
\(\frac{1212}{3131}=\frac{1212:101}{3131:101}=\frac{12}{31}\)
\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)
Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha
\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)