## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises - Page 44: 84

#### Answer

$\dfrac{(7-3x)^{1/2}+\frac{3}{2}x(7-3x)^{-1/2}}{7-3x}=\dfrac{3x-14}{2(3x-7)\sqrt{7-3x}}$

#### Work Step by Step

$\dfrac{(7-3x)^{1/2}+\frac{3}{2}x(7-3x)^{-1/2}}{7-3x}$ Take out common factor $(7-3x)^{-1/2}$ from the numerator: $\dfrac{(7-3x)^{1/2}+\frac{3}{2}x(7-3x)^{-1/2}}{7-3x}=...$ $...=\dfrac{(7-3x)^{-1/2}[(7-3x)+\frac{3}{2}x]}{7-3x}=...$ Simplify the expression inside brackets: $...=\dfrac{(7-3x)^{-1/2}\Big(7-3x+\dfrac{3}{2}x\Big)}{7-3x}=\dfrac{(7-3x)^{-1/2}\Big(7-\dfrac{3}{2}x\Big)}{7-3x}=...$ Simplify the rational expression: $...=(7-3x)^{-3/2}\Big(7-\dfrac{3}{2}x\Big)=\dfrac{7-\dfrac{3}{2}x}{(7-3x)^{3/2}}=\dfrac{14-3x}{2(7-3x)^{3/2}}=...$ $...=\dfrac{14-3x}{2\sqrt{(7-3x)^{3}}}=\dfrac{14-3x}{2(7-3x)\sqrt{7-3x}}=\dfrac{3x-14}{2(3x-7)\sqrt{7-3x}}$

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