a: \(\left(-\dfrac{1}{30}\right)^2+\left(-\dfrac{2}{5}\right)^3\cdot125-\left(-\dfrac{95}{12}\right)^0\)
\(=\dfrac{1}{900}+\dfrac{-8}{125}\cdot125-1\)
\(=\dfrac{1}{900}-9=\dfrac{1-8100}{900}=\dfrac{-8099}{900}\)
b: \(\dfrac{15^3+5\cdot15^2-5^3}{18^3+6\cdot18^2-6^3}=\dfrac{5^3\cdot3^3+5^3\cdot3^2-5^3}{6^3\cdot3^3+6^3\cdot3^2-6^3}\)
\(=\dfrac{5^3\left(3^3+3^2-1\right)}{6^3\left(3^3+3^2-1\right)}=\dfrac{5^3}{6^3}=\dfrac{125}{216}\)
c: \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
=>\(\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
=>2n-1=3
=>2n=4
=>n=2
d: \(-\dfrac{32}{\left(-2\right)^n}=4\)
=>\(\left(-2\right)^n=-\dfrac{32}{4}=-8=\left(-2\right)^3\)
=>n=3
e: \(2^{x+2}-2^x=96\)
=>\(2^x\cdot2^2-2^x=96\)
=>\(3\cdot2^x=96\)
=>\(2^x=\dfrac{96}{3}=32\)
=>x=5
\(a,\left(-\dfrac{1}{30}\right)^2+\left(-\dfrac{2}{5}\right)^3.125-\left(-\dfrac{95}{12}\right)^0\)
\(=\dfrac{1}{900}+\dfrac{-8}{125}.125-1\)
\(=\dfrac{1}{900}-\dfrac{8.1}{1}-1\)
\(=\dfrac{1}{900}-8-1\)
\(=\dfrac{1}{900}-\dfrac{7200}{900}-\dfrac{900}{900}\)
\(=\dfrac{1-7200-900}{900}\)
\(=\dfrac{-8099}{900}\)
\(b,\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)
\(=\dfrac{\left(5.3\right)^3+5.\left(5.3\right)^2-5^3}{\left(6.3\right)^3+6.\left(6.3\right)^2-6^3}\)
\(=\dfrac{5^3.3^3+5.5^2.3^2-5^3}{6^3.3^3+6.6^2.3^2-6^3}\)
\(=\dfrac{5^3.3^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)
\(=\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)
\(=\dfrac{5^3}{6^3}=\dfrac{125}{216}\)
\(c,\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
\(2n-1=3\)
\(2n=3+1\)
\(2n=4\)
\(n=4:2\)
\(n=2\)
Vậy \(n=2\)
\(d,\dfrac{-32}{\left(-2\right)^n}=4\)
\(\left(-2\right)^n=\left(-32\right):4\)
\(\left(-2\right)^n=-8\)
\(\left(-2\right)^n=\left(-2\right)^3\)
\(n=3\)
Vậy \(n=3\)
\(e,2^{x+2}-2^x=96\)
\(2^x.2^2-2^x=96\)
\(2^x.\left(2^2-1\right)=96\)
\(2^x.\left(4-1\right)=96\)
\(2^x.3=96\)
\(2^x=96:3\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
Vậy \(x=5\)
