15.4^12.97-4.3^15.8^8 phần 19.2^24.3^14-6.4^12.27^5
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a) A = 110 - (-761) + 296 + 1454 - (-813 + 1077)
= 110 + 761 + 296 + 1454 - 264
= 871 + 1750 - 264
= 2631 - 264
= 2357
\(B=\frac{15.4^{12}.9^7-4.3^{15}.8^8}{19.2^{24}.3^{14}-6.4^{12}.27^5}=\frac{3}{1}=3\)
\(=\frac{3.5.\left(2^2\right)^{12}.\left(3^2\right)^7-2^2.3^{15}.\left(2^3\right)^8}{19.2^{24}.3^{14}-2.3.\left(2^2\right)^{12}.\left(3^3\right)^5}\)
\(=\frac{3.5.2^{24}.3^{14}-2^2.3^{15}.2^{24}}{19.2^{24}.3^{14}-2.3.2^{24}.3^{15}}\)
\(=\frac{5.2^{24}.3^{15}-3^{15}.2^{26}}{19.2^{24}.3^{14}-2^{25}.3^{16}}\)
\(=\frac{2^{24}.3^{15}.\left(5-2^2\right)}{2^{24}.3^{14}.\left(19-3^2\right)}\)
\(=\frac{3.1}{10}\)
\(=\frac{3}{10}\)
a) \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\frac{18.\left(-130\right)}{\left(-12\right).\left(-130\right)}=\frac{-3}{2}\)
b) \(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{15.12}{12.3}=5\)
a) Ta có: \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\frac{2.\left(-13\right).3^2.2.5}{3.2^2.5.2.13}\)
\(=\frac{\left(-13\right).2^2.3^2.5}{2^3.3.5.13}\)
\(=-\frac{3}{2}\)
b) Ta có: \(\frac{15.8+15.4}{12.3}=\frac{3.5.\left(8+4\right)}{2^2.3.3}\)
\(=\frac{3.5.2^2.3}{2^2.3.3}\)
\(=5\)
\(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{15.12}{12.3}=5\)
a, ta có \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}\)=\(\frac{2.\left(-13\right).\left(-3\right).\left(-3\right).2.\left(-5\right)}{\left(-3\right).2.2.\left(-5\right).\left(-2\right).\left(-13\right)}\)
rút gọn đi còn: \(\frac{-3}{-2}\)=\(\frac{3}{2}\)
Ta có: \(\left(-\dfrac{3}{2}\right)^3+\dfrac{-49}{24}\cdot\dfrac{3}{14}-\left(\dfrac{5}{8}-\dfrac{27}{8}\right)\)
\(=\dfrac{-27}{8}+\dfrac{-7}{2}\cdot\dfrac{1}{8}-\dfrac{-22}{8}\)
\(=\dfrac{-27+22}{8}+\dfrac{-7}{16}\)
\(=\dfrac{-5}{8}+\dfrac{-7}{16}\)
\(=\dfrac{-10}{16}+\dfrac{-7}{16}\)
\(=\dfrac{-17}{16}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\left(8+4\right)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=5\)
Vậy : \(\frac{15\cdot8+15\cdot4}{12\cdot3}=5\)