# Thread: neg. exponents and fractions

1. ## 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.

2. Hello, shiznid12!

Here's the third one . . .

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

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

Get a common denominator:

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

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