Suppose $\displaystyle x$ is nonnegative.

Then $\displaystyle |x| = x$. 

Consequently,

$\displaystyle \left | - 7 \right | \cdot \left | x\right | + \left | -4 \right | = 7 x +4$,

which must be positive,

and 

$\displaystyle (-7) \cdot x+ (-4) = -7x-4$,

which is the opposite of $\displaystyle 7 x +4$ and consequently must be negative. Therefore, (a) is greater.

Suppose $\displaystyle x$ is negative. 

Then $\displaystyle |x| = -x$.

Consequently,

$\displaystyle \left | - 7 \right | \cdot \left | x\right | + \left | -4 \right | = 7( -x) +4 = -7x+4$,

and 

$\displaystyle (-7) \cdot x+ (-4) = -7x-4$.

$\displaystyle 4 > -4$, so

$\displaystyle -7x+4> -7x-4$,

and (a) is greater.

(a) is the greater quantity either way.

$\displaystyle f (x) = 3x$, so $\displaystyle f (2) = 3 \cdot 2 = 6$.

$\displaystyle (g \circ f) (2) = 6$, so, by definition, $\displaystyle g[ f (2) ] = 6$, or $\displaystyle g(6) = 6$.

The graph of $\displaystyle g(x)$ is a line through the point with coordinates $\displaystyle (6,6)$ and with slope $\displaystyle \frac{1}{2}$. The equation of the line can be determined by setting $\displaystyle g(x) = x = 6 , m = \frac{1}{2}$ in the slope-intercept form:

$\displaystyle g(x) = mx+ b$

$\displaystyle 6 = \frac{1}{2} \cdot 6+ b$

$\displaystyle 6= 3+ b$

$\displaystyle b = 3$.

The equation of the line is $\displaystyle g(x) = \frac{1}{2} x+ 3$, which makes this the definition of $\displaystyle g(x)$. By setting $\displaystyle x= 2$,

$\displaystyle g(2) = \frac{1}{2} \cdot 2+ 3 = 1 + 3 = 4$.

Therefore, $\displaystyle f(2)> g(2)$

If $\displaystyle x- y = 125$,

then 

$\displaystyle y - x = - (x-y) = -125$.

The absolute value of a negative number is its (positive) opposite, so

$\displaystyle |y - x |= | -125| = 125$

Also, if $\displaystyle x$ and $\displaystyle y$ are both positive, then $\displaystyle y + x$ is positive; the absolute value of a positive number is the number itself, so $\displaystyle | y + x| = y+ x$. Since $\displaystyle x- y = 125$, it follows that $\displaystyle x = 125 + y$. Therefore, 

$\displaystyle | y + x| = y+ x = y + (125+y ) = 125 + 2y$

Since $\displaystyle y$ is given to be positive,

$\displaystyle 125 + 2y > 125$. 

and

$\displaystyle | y + x| > |y - x|$

Group and combine like terms $\displaystyle 5x^{2} y ^{2},6x^{2}y^{2}$:

$\displaystyle 5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

$\displaystyle = 5x^{2} y ^{2}+ 6x^{2}y^{2}+ 11 xy +7$

$\displaystyle =\left ( 5+ 6 \right )x^{2}y^{2}+ 11 xy +7$

$\displaystyle =11x^{2}y^{2}+ 11 xy +7$

$\displaystyle (x + y)^{2} = x^{2}+2xy + y^{2}$

Since $\displaystyle x$ and $\displaystyle y$ have different signs,

$\displaystyle xy< 0$, and, subsequently,

$\displaystyle 2xy < 0$

Therefore, 

$\displaystyle (x + y)^{2} = x^{2}+2xy + y^{2} < x^{2} + y^{2}$

This makes (b) the greater quantity.

Simplify the expression in (a):

$\displaystyle \frac{(x+y)^{2} + (x - y)^{2}}{x^{2}+y^{2}}$

$\displaystyle =\frac{(x^{2}+2xy + y^{2})+ (x^{2}-2xy + y^{2})}{x^{2}+y^{2}}$

$\displaystyle =\frac{ x^{2}+x^{2} +2xy -2xy+ y^{2}+ y^{2}}{x^{2}+y^{2}}$

$\displaystyle =\frac{ 2x^{2} +2 y^{2}}{x^{2}+y^{2}} =\frac{ 2(x^{2}+y^{2})}{x^{2}+y^{2}} = 2$

Therefore, whether (a) or (b) is greater depends on the values of $\displaystyle x$ and $\displaystyle y$, neither of which are known. 

We give at least one positive value of $\displaystyle x$ for which (a) is greater and at least one positive value of $\displaystyle x$ for which (b) is greater.

Case 1: $\displaystyle x = 2$

(a) $\displaystyle x^{2} + 3x = 2^{2} + 3 \cdot 2 = 4 + 6 = 10$

(b) $\displaystyle 4x = 4 \cdot 2 = 8$

Case 2: $\displaystyle x = \frac{1}{2}$

(a) $\displaystyle x^{2} + 3x = \left ( \frac{1}{2} \right ) ^{2} + 3 \cdot \frac{1}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}= 1 \frac{3}{4}$

(b) $\displaystyle 4x = 4 \cdot \frac{1}{2} = 2$

Therefore, either (a) or (b) can be greater.

Any nonzero expression raised to the power of 0 is equal to 1. Therefore, 

$\displaystyle \left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0} = 1 + 1 = 2$.

None of the given expressions are correct.

$\displaystyle \left (\frac{x^{3}}{2y^{4}} \right ) ^{-3} = \left (\frac{2y^{4}}{x^{3}} \right ) ^{3} = \frac{2^{3} \left (y^{4}\right ) ^{3} }{\left (x^{3} \right )^{3}}= \frac{8 y^{4 \cdot 3} }{x^{3\cdot 3} }= \frac{8 y^{12} }{x^{9} }$

If $\displaystyle y > 1$, then $\displaystyle y ^{2}> 1^{2} = 1$ and $\displaystyle y ^{-2}= \frac{1}{y ^{2}} < \frac{1}{1} = 1$

$\displaystyle y ^{2}> 1> y ^{-2}$, so by transitivity, $\displaystyle y ^{2}> y ^{-2}$, and (b) is greater