Question

1. simplify the expression:

\frac{(-x^{0}y^{-5})(2x)^{3}}{4xy^{2}}

Explanation:

Step1: Simplify exponents and powers

First, recall that \(a^0 = 1\) (for \(a
eq0\)) and \((ab)^n=a^n b^n\). So, \(x^0 = 1\) and \((2x)^3=2^3x^3 = 8x^3\). The numerator becomes \((-1\times1\times y^{-5})(8x^3)=-8x^3y^{-5}\). Now the expression is \(\frac{-8x^3y^{-5}}{4xy^2}\).

Step2: Simplify coefficients and variables

For the coefficients: \(\frac{-8}{4}=-2\). For the \(x\)-terms: use the rule \(\frac{x^m}{x^n}=x^{m - n}\), so \(x^3\div x = x^{3 - 1}=x^2\). For the \(y\)-terms: \(y^{-5}\div y^2=y^{-5 - 2}=y^{-7}\), and \(y^{-7}=\frac{1}{y^7}\). Combining these, we get \(-2\times x^2\times\frac{1}{y^7}=\frac{-2x^2}{y^7}\).

Answer:

\(\boldsymbol{\frac{-2x^2}{y^7}}\)