Question

Simplify each expression. Write your final answer without negative exponents.

(a)$\sqrt{200}-\sqrt{32}$

(b)$\left(3a^3{b}^3\right){\left(4a{b}^2\right)}^2$

(c)${\left(\frac{3x^{\frac{3}{2}}{y}^3}{x^2{y}^{-\frac{1}{2}}}\right)}^{-2}$

(d)$\frac{x^2+3x+2}{x^2-x-2}$

(e)$\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$

(f)$\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}$

Short answer

(a)$6\sqrt{2}$

(b)$48a^5{b}^7$

(c)$\frac{x}{9y^7}$

(d)$\frac{x+2}{x-2}$

(e)$\frac{1}{x^2-4}$

(f)$-(y+x)$

Step-by-step solution

Part a. Step 1. Concept of radicals.

The symbol ‘$\sqrt{}$‘ means ‘the positive square root of”

Thus

$\sqrt{a}=b$means $b^2=a\mathrm{and}b\ge 0$

Square root are special case of$\mathrm{nth}\mathrm{roots}=\mathrm{The}\mathrm{nth}\mathrm{roots}\mathrm{of}x$ is the number that, when raised to the$\mathrm{nth}$ power given x

Part a. Step 2. Given expression.

The expression is$\sqrt{200}-\sqrt{32}$

Part a. Step 3. Solving the expression.

$\begin{array}{l}\sqrt{200}-\sqrt{32}\\ \sqrt{100.2}-\sqrt{16.2}(\mathrm{Factor}\mathrm{out}\mathrm{the}\mathrm{largest}\mathrm{squares})\\ \sqrt{100}\sqrt{2}-\sqrt{16}\sqrt{2}(\sqrt{ab}=\sqrt{a}\sqrt{b})\\ 10\sqrt{2}-4\sqrt{2}(\mathrm{Distributive}\mathrm{property})\\ 6\sqrt{2}\end{array}$

Part b. Step 1. Concept of working exponent.

${\left(ab\right)}^n=a^n{b}^n$

And,

${\left(a^m\right)}^n=a^{mn}$

And$a^m{a}^n=a^{m+n}$

Part b. Step 2. Given expression.

The expression is$\left(3a^3{b}^3\right){\left(4a{b}^2\right)}^2$

Part b. Step 3. Solving the expression.

$$\begin{gathered} \left(3a^3{b}^3\right){\left(4a{b}^2\right)}^2=\left(3a^3{b}^3\right)\left[4^2{a}^2{\left(b^2\right)}^2\right]\left(\right(a{b}^n)=a^n{b}^n) \\ =\left(3a^3{b}^3\right)\left(16a^2{b}^4\right)\left[{\left(a^m\right)}^n=a^{mn}\right] \\ =\left(3\right)\left(16\right)a^3{a}^2{b}^3{b}^4(\mathrm{Group}\mathrm{factors}\mathrm{with}\mathrm{the}\mathrm{same}\mathrm{base}) \\ =48{\mathrm{a}}^5{\mathrm{b}}^7[{\mathrm{a}}^{\mathrm{m}}{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}] \end{gathered}$$

Part c. Step 1. Concept of working exponent.

${\left(ab\right)}^n=a^n{b}^n$

And,

${\left(a^m\right)}^n=a^{mn}$

And$a^m{a}^n=a^{m+n}$

Part c. Step 2. Given expression.

The expression is${\left(\frac{3x^{\frac{3}{2}}{y}^3}{x^2{y}^{-\frac{1}{2}}}\right)}^{-2}$

Part c. Step 3. Solving the expression.

$$\begin{gathered} {\left(\frac{3x^{\frac{3}{2}}{y}^3}{x^2{y}^{-\frac{1}{2}}}\right)}^{-2}={\left(\frac{x^2{y}^{-\frac{1}{2}}}{3x^{\frac{3}{2}}{y}^3}\right)}^2 \\ =\frac{{\left(x^2\right)}^2{\left(y^{-\frac{1}{2}}\right)}^2}{3^2{\left(x^{\frac{3}{2}}\right)}^2{\left(y^3\right)}^2} \\ =\frac{x^4{y}^{-1}}{9x^3{y}^6} \\ =\frac{x^4}{9x^3{y}^7} \\ =\frac{x}{9y^7} \end{gathered}$$

Part d. Step 1. Concept of working exponent.

${\left(ab\right)}^n=a^n{b}^n$

And,

${\left(a^m\right)}^n=a^{mn}$

And$a^m{a}^n=a^{m+n}$

Part d. Step 2. Given expression.

The expression is$\frac{x^2+3x+2}{x^2-x-2}$

Part d. Step 3. Solving the expression.

$$\begin{gathered} \frac{x^2+3x+2}{x^2-x-2}=\frac{(x+1)(x+2)}{(x-2)(x+1)}\left(\mathrm{Factor}\right) \\ =\frac{x+2}{x-2} \end{gathered}$$

Part e. Step 1. Concept of working exponent.

${\left(ab\right)}^n=a^n{b}^n$

And,

${\left(a^m\right)}^n=a^{mn}$

And$a^m{a}^n=a^{m+n}$

Part e. Step 2. Given expression.

The expression is$\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$

Part e. Step 3. Solving the expression.

$$\begin{gathered} \frac{x^2}{x^2-4}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}\left(\mathrm{Factor}\right) \\ =\frac{x^2-(x+1)(x-1)}{(x-2)(x+2)}\left(\mathrm{Taking}\mathrm{LCMand}\mathrm{solving}\right) \\ =\frac{x^2-x^2+1}{(x-2)(x+2)}\left(\mathrm{Simplify}\right) \\ =\frac{1}{x^2-4} \end{gathered}$$

Part f. Step 1. Concept of working exponent.

${\left(ab\right)}^n=a^n{b}^n$

And,

${\left(a^m\right)}^n=a^{mn}$

And$a^m{a}^n=a^{m+n}$

Part f. Step 2. Given expression.

The expression is$\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}$

Part f. Step 3. Solving the expression.

$$\begin{gathered} \frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}=\frac{\frac{y^2-x^2}{xy}}{\frac{x-y}{xy}}(\mathrm{Taking}\mathrm{LCM}\mathrm{and}\mathrm{solving}) \\ =\frac{y^2-x^2}{x-y}\left(\mathrm{Simplify}\right) \\ =\frac{(y-x)(y+x)}{(x-y)}\left(\mathrm{Factor}\right) \\ =-(y+x) \end{gathered}$$