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\(=\dfrac{7^{16}\cdot2^{16}\cdot7^{32}\cdot3^{32}\cdot7^{48}\cdot5^{48}}{5^{16}\cdot2^{16}\cdot5^{32}\cdot3^{32}\cdot7^{96}}\)
\(=\dfrac{7^{96}\cdot2^{16}\cdot3^{32}\cdot5^{48}}{7^{96}\cdot2^{48}\cdot5^{48}}=\dfrac{3^{32}}{2^{32}}=1.5^{32}\)
\(\frac{14^{16}\cdot21^{31}\cdot35^{48}}{10^{16}\cdot15^{32}\cdot7^{96}}\)
\(=\frac{\left(2\cdot7\right)^{16}\cdot\left(3\cdot7\right)^{31}\cdot\left(5\cdot7\right)^{48}}{\left(2\cdot5\right)^{16}\cdot\left(3\cdot5\right)^{32}\cdot\left(7^2\right)^{48}}\)
\(=\frac{2^{16}\times3^{31}\times5^{48}\times7^{95}}{2^{16}\times3^{32}\times5^{48}\times7^{96}}\)
\(=\frac{1\times1}{3\times7}\)
\(=\frac{1}{21}\)
a: \(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1-\dfrac{15}{17}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\left(\dfrac{12}{17}-1-\dfrac{15}{17}\right)+1\)
\(=\dfrac{-20}{17}+1=\dfrac{-3}{17}\)
b: \(B=\dfrac{-5}{3}\cdot16\dfrac{2}{7}-\dfrac{-5}{3}\cdot28\dfrac{2}{7}\)
\(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}-28-\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(-12\right)=20\)
c: \(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-25}{27}+\dfrac{1}{5}-1\)
\(=\dfrac{-125+27-135}{135}=\dfrac{-233}{135}\)
Bài làm
\(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(C=\frac{2.5^{21}.5-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{14}.7-19.7^{14}\right)}{7^{16}.7+3.7^{15}}\)
\(C=\frac{5^{21}.\left(2-9\right).5}{\left(5.5\right)^{10}}:\frac{5.[7^{15}.\left(3-19\right)].7}{7^{15}.\left(3+1\right).7}\)
\(C=\frac{5^{21}.\left(-7\right).5}{5^{10}.5^{10}}:\frac{5.7^{15}.\left(-16\right).7}{7^{15}.4.7}\)
\(C=\frac{5^{21}.\left(-35\right)}{5^{10}.5^{10}}:\frac{7^{15}.\left(-112\right).5}{7^{15}.28}\)
\(C=5.\left(-35\right):\frac{7^{15}.560}{7^{15}.28}\)
\(C=5.\left(-35\right):\frac{1.560}{1.28}\)
\(C=5.\left(-35\right):20\)
\(C=5.\left(-35\right).\frac{1}{20}\)
\(C=-\frac{175}{20}\)
\(C=-\frac{35}{4}\)
Vậy biểu thức trên \(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)bằng \(C=-\frac{35}{4}\)
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