Question

1. \\(\frac{m^4}{2m^4}\\)

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

1. \\(\frac{2h^3 j^{-3} k^4}{3jk}\\)

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

Explanation:

Problem 24: $\boldsymbol{\frac{m^4}{2m^4}}$

Step1: Cancel common terms

The numerator and denominator both have $m^4$, so we can cancel them out.
$\frac{m^4}{2m^4} = \frac{1}{2} \cdot \frac{m^4}{m^4}$

Step2: Simplify the fraction

Since $\frac{m^4}{m^4} = 1$ (for $m
eq 0$), we have $\frac{1}{2} \cdot 1 = \frac{1}{2}$

Step1: Use exponent rules for division ($a^m / a^n = a^{m - n}$)

For $x$: $x^{4 - 2} = x^2$
For $y$: $y^{-4 - (-3)} = y^{-4 + 3} = y^{-1}$
For $z$: $z^{-3 - 4} = z^{-7}$
The coefficient remains $\frac{2}{3}$.
So the expression becomes $\frac{2}{3}x^2y^{-1}z^{-7}$

Step2: Rewrite negative exponents (optional, for positive exponents)

$y^{-1} = \frac{1}{y}$ and $z^{-7} = \frac{1}{z^7}$, so we can write it as $\frac{2x^2}{3yz^7}$

Step1: Apply exponent rules for division

For $j$: $j^{-3 - 1} = j^{-4}$
For $k$: $k^{4 - 1} = k^3$
The coefficient is $\frac{2}{3}$ and $h^3$ remains.
So the expression becomes $\frac{2}{3}h^3j^{-4}k^3$

Step2: Rewrite negative exponent (optional)

$j^{-4} = \frac{1}{j^4}$, so we get $\frac{2h^3k^3}{3j^4}$

Answer:

$\frac{1}{2}$

Problem 26: $\boldsymbol{\frac{2x^4y^{-4}z^{-3}}{3x^2y^{-3}z^4}}$