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1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
\(\dfrac{2^{15}.9^3}{6^7.4^4}\\ =\dfrac{2^{15}.3^9}{3^7.2^7.2^8}\\ =\dfrac{2^{15}.3^9}{3^7.2^{15}}\\ =\dfrac{3^9}{3^7}\\ =3^2\\ =9\)
\(\dfrac{2^{15}.9^3}{6^7.4^4}=\dfrac{2^{15}.\left(3^2\right)^3}{3^7.2^7.2^8}=\dfrac{2^{15}.3^6}{3^7.2^{15}}=\dfrac{3^6}{3^7}=\dfrac{1}{3}\)
\(\dfrac{2^{15}.9^3}{6^7.4^4}=\dfrac{2^{15}.3^6}{2^{15}.3^7}=\dfrac{1}{3}\)
a) \(\frac{-5}{8}\cdot\frac{11}{3}+\frac{-5}{8}\cdot\frac{1}{3}=-\frac{5}{8}\left(\frac{11}{3}+\frac{1}{3}\right)=-\frac{5}{8}\cdot4=-\frac{5}{2}\cdot1=-\frac{5}{2}\)
b) \(\frac{2}{3}+\frac{3}{4}\cdot\frac{9}{5}=\frac{2}{3}+\frac{27}{20}=\frac{121}{60}\)
c) Tương tự câu a
d) \(\frac{1}{7}\cdot\frac{3}{8}+\frac{1}{7}\cdot\frac{5}{8}=\frac{1}{7}\left(\frac{3}{8}+\frac{5}{8}\right)=\frac{1}{7}\cdot1=\frac{1}{7}\)
\(a,\frac{-5}{8}.\frac{11}{3}+\frac{-5}{8}.\frac{1}{3}\)
\(=\frac{-5}{8}\left(\frac{11}{3}+\frac{1}{3}\right)\)
\(=\frac{-5}{8}.4\)
\(=\frac{-5}{2}\)
\(b,\frac{2}{3}+\frac{3}{4}.\frac{9}{5}\)
\(=\frac{2}{3}+\frac{27}{20}\)
\(=\frac{40}{60}+\frac{81}{60}\)
\(=\frac{121}{60}\)
\(c,\frac{-5}{7}.\frac{4}{9}-\frac{5}{9}.\frac{5}{7}\)
\(=\frac{-5}{7}\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=\frac{-5}{7}.1\)
\(=\frac{-5}{7}\)
\(d,\frac{1}{7}.\frac{3}{8}+\frac{1}{7}.\frac{5}{8}\)
\(=\frac{1}{7}\left(\frac{3}{8}+\frac{5}{8}\right)\)
\(=\frac{1}{7}.1\)
\(=\frac{1}{7}\)
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