# neg. exponents and fractions

• Feb 1st 2010, 06:23 AM
shiznid12
neg. exponents and fractions
There are 3 problems I am having issues with due to order of operations and negative exponents/fractions:

(2x(x + 6)^4) − (4x^2(x + 6)^3)

(x^-1(2 − x^2)) + (x^-2(2 − x^2)^2)

(7-3x)^(2/3) + 3/2x(7-3x)^-(1/3)

Help of any sort would be nice. Solutions are cool so I can work through it to figure out how to get the answer, or hints on steps to take are awesome too.

Appreciate it, guys, you are awesome. :)
• Feb 1st 2010, 08:41 AM
Soroban
Hello, shiznid12!

Here's the third one . . .

Quote:

$(7-3x)^{\frac{2}{3}} + \tfrac{3}{2}x(7-3x)^{-\frac{1}{3}}$

We have: . $(7-3x)^{\frac{2}{3}} + \frac{3x}{2(7-3x)^{\frac{1}{3}}}$

Get a common denominator:

. . $\frac{(7-3x)^{\frac{2}{3}}}{1}\cdot{\color{blue}\frac{2(7-3x)^{\frac{1}{3}}}{2(7-3x)^{\frac{1}{3}}}} \:+\: \frac{3x}{2(7-3x)^{\frac{1}{3}}}$ . $=\;\;\frac{2(7-3x)}{2(7-3x)^{\frac{1}{3}}} + \frac{3x}{2(7-3x)^{\frac{1}{3}}}$

. . . $=\;\; \frac{2(7-3x) + 3x}{2(7-3x)^{\frac{1}{3}}} \;\;=\;\;\frac{14 - 6x + 3x}{2(7-3x)^{\frac{1}{3}}} \;\;=\;\;\boxed{ \frac{14-3x}{2(7-3x)^{\frac{1}{3}}}}$