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Đặt giá trị biểu thức là A, ta có:
\(A=\dfrac{\left(-27\right)^{10}.16^{25}}{6^{30}.\left(-32\right)^{15}}\)
\(A=\dfrac{\left(\left(-3\right)^3\right)^{10}.\left(2^4\right)^{25}}{2^{30}.3^{30}.\left(\left(-2\right)^5\right)^{15}}\)
\(A=\dfrac{\left(-3\right)^{30}.2^{100}}{2^{30}.3^{30}.\left(-2\right)^{75}}\)
\(A=\dfrac{3^{30}.2^{100}}{2^{30}.2^{75}.3^{30}}=\dfrac{1}{2^5}=\dfrac{1}{32}\)
a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)
b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)
c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)
a) \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=-\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=\dfrac{-3}{5}\)
b: \(\dfrac{4^{30}\cdot3^{43}}{2^{57}\cdot27^{15}}\)
\(=\dfrac{2^{60}\cdot3^{43}}{2^{57}\cdot3^{45}}\)
\(=8\cdot\dfrac{1}{9}=\dfrac{8}{9}\)
\(\frac{\left(-27\right)^{10}.16^{25}}{-3.\left(-32\right)^{15}}=\frac{\left(-3\right)^{30}.\left(-2\right)^{100}}{\left(-3\right).\left(-2\right)^{75}}=\frac{\left(-3\right).\left(-3\right)^{29}.\left(-2\right)^{75}.\left(-2\right)^{25}}{\left(-3\right).\left(-2\right)^{75}}\)
= \(\left(-3\right)^{29}.\left(-2\right)^{25}\)
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
\(=\dfrac{3^{30}\cdot2^{100}}{2^{30}\cdot3^{30}\cdot2^{75}}=\dfrac{1}{2^5}=\dfrac{1}{32}\)
cảm ơn ạ