[latex]\begin{align} \frac{5a^2}{14}\cdot \frac{7}{10a^3} &= \frac{5a^2\cdot 7}{14\cdot 10a^3} && {\color{blue}\textsf{multiply the numerators and denominators accordingly}}\\[5pt] &= \frac{35a^2}{140a^3} && {\color{blue}\textsf{simplify the numerator and denominator}}\\[5pt] &= \frac{35 \cdot a^2}{35 \cdot 4 \cdot a^2\cdot a} && {\color{blue}\textsf{factor the numerator and denominator; look for the GCF}}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{35}}}\cdot {\color{red}\cancel{\color{black}{a^2}}}\cdot 1}{{\color{red}\cancel{\color{black}{35}}} \cdot 4\cdot {\color{red}\cancel{\color{black}{a^2}}}\cdot a} && {\color{blue}\textsf{divide out the GCF between the numerator and denominator}}\\[5pt] &= \frac{1}{4a} && {\color{blue}\textsf{simplify}}\\[5pt] \end{align}[/latex]

[latex]\begin{align} \frac{5a^2}{14}\cdot \frac{7}{10a^3} &= \frac{5\cdot a^2}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot a^2 \cdot a} && {\color{blue}\textsf{factor the numerators and denominators; look for the GCFs}}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{5}}}\cdot {\color{red}\cancel{\color{black}{a^2}}}\cdot 1}{{\color{red}\cancel{\color{black}{7}}}\cdot 2}\cdot \frac{{\color{red}\cancel{\color{black}{7}}}\cdot 1}{{\color{red}\cancel{\color{black}{5}}}\cdot 2\cdot {\color{red}\cancel{\color{black}{a^2}}} \cdot a} && {\color{blue}\textsf{divide out common factors}}\\[5pt] &= \frac{1}{4a} && {\color{blue}\textsf{simplify}} \end{align}[/latex]

Since $$a=0$$ makes the denominator of the first rational expression equal to $$0$$ and $$a=-1$$ makes the denominator of the second rational expression equal to $$0$$, the restricted values of $$a$$ are $$0$$ and $$-1$$. Now,

[latex]\begin{align} \frac{a^2-a-2}{5a}\cdot \frac{10a}{a+1} &= \frac{(a-2)\cdot (a+1)}{5\cdot a}\cdot \frac{5\cdot 2 \cdot a}{(a+1)} && {\color{blue}\textsf{factor the numerators and denominators}}\\[5pt] &= \frac{(a-2)\cdot {\color{red}\cancel{\color{black}{(a+1)}}}}{{\color{red}\cancel{\color{black}{5}}}\cdot {\color{red}\cancel{\color{black}{a}}}\cdot 1}\cdot \frac{{\color{red}\cancel{\color{black}{5}}}\cdot 2 \cdot {\color{red}\cancel{\color{black}{a}}}}{{\color{red}\cancel{\color{black}{(a+1)}}}\cdot 1} && {\color{blue}\textsf{divide out common factors}}\\[5pt] &= \frac{2(a-2)}{1} && {\color{blue}\textsf{multiply}}\\[5pt] &= 2(a-2) && {\color{blue}\textsf{simplify}} \end{align}[/latex]

The final answer may also be written in expanded (distributed) form [latex]2a-4[/latex].

Since $$2t^2-t-10=(2t-5)(t+2)$$, using the Zero-Product Property, we acknowledge that $$t=\dfrac{5}{2}$$ and $$t=-2$$ make the denominator of the first rational expression equal to $$0$$. Also, $$t^2+2t=t(t+2)$$, so $$t=0$$ and $$t=-2$$ make the denominator of the second rational expression equal to $$0$$. Hence, the restricted values of $$t$$ are $$\dfrac{5}{2}$$, $$0$$, and $$-2$$. Now,

[latex]\begin{align} \frac{t^2+4t+4}{2t^2-t-10}\cdot\frac{t+5}{t^2+2t} &= \frac{(t+2)\cdot (t+2)}{(2t-5)\cdot (t+2)}\cdot\frac{(t+5)}{t\cdot (t+2)} && {\color{blue}\textsf{factor the numerators and denominators}}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{(t+2)}}}\cdot {\color{red}\cancel{\color{black}{(t+2)}}}\cdot1}{(2t-5)\cdot {\color{red}\cancel{\color{black}{(t+2)}}}}\cdot\frac{(t+5)}{t\cdot {\color{red}\cancel{\color{black}{(t+2)}}}} && {\color{blue}\textsf{divide out common factors}}\\[5pt] &= \frac{t+5}{t(2t-5)} && {\color{blue}\textsf{multiply and simplify}} \end{align}[/latex]

The final answer may also be written in expanded (distributed) form [latex]\dfrac{t+5}{2t^2-5t}[/latex].

For the restricted values of $$x$$, first notice the two denominators, [latex]9[/latex] and [latex]27[/latex]. These never equal zero. Since we divide by [latex]\dfrac{15x^3}{27}[/latex], the numerator of that rational expression cannot equal zero either. Hence, $$x\ne 0$$ and the only restricted value of $$x$$ is $$0$$. Now,

[latex]\begin{align} \frac{5x^2}{9}\div\frac{15x^3}{27} &= \frac{5x^2}{9}\cdot\frac{27}{15x^3} && {\color{blue}\textsf{rewrite division as multiplication by the reciprocal}}\\[5pt] &= \frac{5\cdot x^2}{9}\cdot\frac{9\cdot 3}{5\cdot 3 \cdot x^2\cdot x} && {\color{blue}\textsf{factor the numerators and denominators; look for the GCFs}}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{5}}}\cdot {\color{red}\cancel{\color{black}{x^2}}}\cdot 1}{{\color{red}\cancel{\color{black}{9}}}\cdot 1}\cdot\frac{{\color{red}\cancel{\color{black}{9}}}\cdot {\color{red}\cancel{\color{black}{3}}}\cdot 1}{{\color{red}\cancel{\color{black}{5}}}\cdot {\color{red}\cancel{\color{black}{3}}} \cdot {\color{red}\cancel{\color{black}{x^2}}}\cdot x} && {\color{blue}\textsf{divide out common factors}}\\[5pt] &= \frac{1}{x} && {\color{blue}\textsf{multiply and simplify}} \end{align}[/latex]

For the restricted values of $$x$$, first notice that $$x=-2$$ makes the denominator of the first rational expression, $$x+2$$, equal to zero. Since $$x^2+5x+6=(x+3)(x+2)$$, using the Zero-Product Property, we acknowledge that $$x=-3$$ and $$x=-2$$ make the denominator of the second rational expression equal to zero. Since we divide by [latex]\dfrac{6x^4}{x^2+5x+6}[/latex], the numerator of that rational expression, $$6x^4$$, cannot equal zero either. Hence, $$x\ne 0$$. Summarizing, the restricted values of $$x$$ are $$-3$$, $$-2$$, and $$0$$. Now,

[latex]\begin{align} \frac{3x^2}{x+2}\div\frac{6x^4}{x^2+5x+6} &= \frac{3x^2}{x+2}\cdot\frac{x^2+5x+6}{6x^4} && {\color{blue}\textsf{rewrite division as multiplication by the reciprocal}}\\[5pt] &= \frac{3\cdot x^2}{(x+2)}\cdot\frac{(x+3)\cdot (x+2)}{3\cdot 2\cdot x^2\cdot x^2} && {\color{blue}\textsf{factor the numerators and denominators; look for the GCFs}}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{3}}}\cdot {\color{red}\cancel{\color{black}{x^2}}}\cdot 1}{{\color{red}\cancel{\color{black}{(x+2)}}}\cdot 1}\cdot\frac{(x+3)\cdot {\color{red}\cancel{\color{black}{(x+2)}}}}{{\color{red}\cancel{\color{black}{3}}}\cdot 2\cdot {\color{red}\cancel{\color{black}{x^2}}}\cdot x^2} && {\color{blue}\textsf{divide out common factors}}\\[5pt] &= \frac{x+3}{2x^2} && {\color{blue}\textsf{multiply and simplify}} \end{align}[/latex]