Question

1. (\frac{bx}{x^3y^{-2}} - 3by^2x^{-2} + \frac{4y^2}{bx^2}) 2)

Explanation:

Step1: Simplify each term using exponent rules

Recall that \(a^{-n}=\frac{1}{a^n}\) and \(\frac{1}{a^{-n}} = a^n\). For the first term \(\frac{bx}{x^{3}y^{-2}}\), using the rule \(\frac{a^m}{a^n}=a^{m - n}\) for \(x\) and \(y^{-2}=\frac{1}{y^{-2}} = y^{2}\), we have:
\(\frac{bx}{x^{3}y^{-2}}=bx\cdot x^{-3}\cdot y^{2}=b x^{1-3}y^{2}=b x^{-2}y^{2}=\frac{b y^{2}}{x^{2}}\)
The second term is \(-3by^{2}x^{-2}=-\frac{3by^{2}}{x^{2}}\) (since \(x^{-2}=\frac{1}{x^{2}}\))
The third term is \(\frac{4y^{2}}{bx^{2}}\)

Step2: Combine the first two terms

Now, add the first two simplified terms: \(\frac{b y^{2}}{x^{2}}-\frac{3by^{2}}{x^{2}}=\frac{b y^{2}- 3by^{2}}{x^{2}}=\frac{-2by^{2}}{x^{2}}\)

Step3: Add the third term

Now, add the third term \(\frac{4y^{2}}{bx^{2}}\) to the result from Step 2. To add these two fractions, we need a common denominator, which is \(bx^{2}\). So we rewrite \(\frac{-2by^{2}}{x^{2}}\) as \(\frac{-2b^{2}y^{2}}{bx^{2}}\) (multiplying numerator and denominator by \(b\))
Then, \(\frac{-2b^{2}y^{2}}{bx^{2}}+\frac{4y^{2}}{bx^{2}}=\frac{-2b^{2}y^{2}+4y^{2}}{bx^{2}}=\frac{y^{2}(-2b^{2} + 4)}{bx^{2}}=\frac{2y^{2}(2 - b^{2})}{bx^{2}}\) (factoring out 2 from the numerator)

Answer:

\(\frac{2y^{2}(2 - b^{2})}{bx^{2}}\) (or equivalent forms like \(\frac{-2b^{2}y^{2}+4y^{2}}{bx^{2}}\))