Tính rồi rút gọn:
\(\frac{2}{3}\)+ \(\frac{3}{9}\)+ \(\frac{7}{5}\)
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\(F=\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^9\div\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
\(F=\frac{\left(\frac{2.5}{5}\right)^7+\left(\frac{9.16}{4.3}\right)^3}{2^7.5^2+2^9}=\frac{2^7+12^3}{2^7.5^2+2^9}=\frac{2^7+2^6.3^3}{2^7.5^2+2^9}=\frac{2^6.\left(2+3^3\right)}{2^7.\left(5^2+2^2\right)}=\frac{2^6.29}{2^7.29}\)
\(F=\frac{1}{2}\)
1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=\frac{-2x}{3}+3x\left(\frac{15x+20-126}{90}\right)-\frac{5x^2-20x}{10}\)
\(M=\frac{-2x}{3}+3x.\frac{15x-106}{90}-\frac{5.\left(x^2-4x\right)}{10}\)
\(M=\frac{-2x}{3}+\frac{45x^2-318x}{90}-\frac{x^2-4x}{2}\)
\(C=\frac{\left(\frac{2}{5}\right)^7\times5^7+\left(\frac{9}{4}\right)^3\div\left(\frac{3}{16}\right)^3}{2^7\times5^2+512}\)
\(=\frac{\left(\frac{2}{5}\times5\right)^7+\left(\frac{9}{4}\div\frac{3}{16}\right)^3}{2^7\times5^2+2^9}\)
\(=\frac{2^7+12^3}{2^7\times\left(25+2^2\right)}\)
\(=\frac{2^7+\left(2^2\times3\right)^3}{2^7\times\left(25+4\right)}\)
\(=\frac{2^7+2^6\times3^3}{2^7\times29}\)
\(=\frac{2^6\times\left(2+27\right)}{2^7\times29}\)
\(=\frac{29}{2\times29}\)
\(=\frac{1}{2}\)
\(\frac{2}{3}+\frac{3}{9}+\frac{7}{5}\)
\(=\frac{2}{3}+\frac{1}{3}+\frac{7}{5}\)
\(=\frac{10}{15}+\frac{5}{15}+\frac{21}{15}\)
\(=\frac{36}{15}\)
\(=\frac{12}{5}\)
\(\frac{2}{3}+\frac{3}{9}+\frac{7}{5}\)
=\(\frac{2}{3}+\frac{1}{3}+\frac{7}{5}\)
=\(\frac{7}{5}+\frac{2}{3}+\frac{1}{3}\)
=\(\frac{3}{3}+\frac{7}{5}\)
\(=1+\frac{7}{5}\)
=\(\frac{12}{5}\)