Question

1. \\(\frac{2y}{x^2 y^3}\\)
2. \\(\frac{4x^{-1} y^5}{-3x^0}\\)

Explanation:

Step1: Simplify Problem 3

We have the expression \(\frac{2y}{x^{2}y^{3}}\). Using the rule of exponents \(\frac{a^{m}}{a^{n}} = a^{m - n}\) and \(\frac{1}{a^{n}}=a^{-n}\), we can simplify the \(y\) terms and the \(x\) term. For the \(y\) terms: \(\frac{y}{y^{3}}=y^{1 - 3}=y^{-2}=\frac{1}{y^{2}}\). The \(x\) term remains as \(x^{-2}\) (since \(\frac{1}{x^{2}} = x^{-2}\)). So the simplified form is \(\frac{2}{x^{2}y^{2}}\) or \(2x^{-2}y^{-2}\).

Step2: Simplify Problem 6

We have the expression \(\frac{4x^{-1}y^{5}}{-3x^{0}}\). First, recall that \(x^{0}=1\) for \(x
eq0\). Then, using the rule of exponents \(\frac{x^{m}}{x^{n}}=x^{m - n}\), we can simplify the \(x\) terms: \(\frac{x^{-1}}{x^{0}}=x^{-1-0}=x^{-1}=\frac{1}{x}\). Then, the coefficient is \(\frac{4}{-3}=-\frac{4}{3}\). So the simplified form is \(-\frac{4}{3}x^{-1}y^{5}\) or \(-\frac{4y^{5}}{3x}\).

Answer:

Problem 3: \(\boldsymbol{\frac{2}{x^{2}y^{2}}}\) (or \(2x^{-2}y^{-2}\))
Problem 6: \(\boldsymbol{-\frac{4y^{5}}{3x}}\) (or \(-\frac{4}{3}x^{-1}y^{5}\))