$$ \begin{array}{ll} ext { Worksheet } 2 B \ ext { 1. } & \left(a+\frac{1}{a} ight)^{2} \ ext { 2. } & \left(2 a+\frac{1}{2} ight)^{2} \ ext { 3. } & \left(\frac{1}{x}-3 ight)\left(\frac{3}{x}+1 ight) \ ext { 4. } & \left(\frac{2}{x}-3 ight)\left(\frac{2}{x}+3 ight) \ ext { 5. } & \left(\frac{1}{3} x-\frac{3}{4} y ight)^{2}\end{array} $$

Question

$$ \begin{array}{ll}	ext { Worksheet } 2 B \ 	ext { 1. } & \left(a+\frac{1}{a}ight)^{2} \ 	ext { 2. } & \left(2 a+\frac{1}{2}ight)^{2} \ 	ext { 3. } & \left(\frac{1}{x}-3ight)\left(\frac{3}{x}+1ight) \ 	ext { 4. } & \left(\frac{2}{x}-3ight)\left(\frac{2}{x}+3ight) \ 	ext { 5. } & \left(\frac{1}{3} x-\frac{3}{4} yight)^{2}\end{array} $$

UpStudy AI · Accepted Answer

Sure! Let's expand each of the given expressions step by step.

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**1.** $$ \left(a + \frac{1}{a}\right)^{2} $$

**Expansion:**
$$
\left(a + \frac{1}{a}\right)^{2} = a^{2} + 2 \cdot a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^{2} = a^{2} + 2 + \frac{1}{a^{2}}
$$

**Simplified Form:**
$$
a^{2} + 2 + \frac{1}{a^{2}}
$$

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**2.** $$ \left(2a + \frac{1}{2}\right)^{2} $$

**Expansion:**
$$
\left(2a + \frac{1}{2}\right)^{2} = (2a)^{2} + 2 \cdot 2a \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^{2} = 4a^{2} + 2a + \frac{1}{4}
$$

**Simplified Form:**
$$
4a^{2} + 2a + \frac{1}{4}
$$

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**3.** $$ \left(\frac{1}{x} - 3\right)\left(\frac{3}{x} + 1\right) $$

**Expansion Using FOIL:**
$$
\left(\frac{1}{x} - 3\right)\left(\frac{3}{x} + 1\right) = \frac{1}{x} \cdot \frac{3}{x} + \frac{1}{x} \cdot 1 - 3 \cdot \frac{3}{x} - 3 \cdot 1
$$
$$
= \frac{3}{x^{2}} + \frac{1}{x} - \frac{9}{x} - 3
$$

**Combine Like Terms:**
$$
\frac{3}{x^{2}} - \frac{8}{x} - 3
$$

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**4.** $$ \left(\frac{2}{x} - 3\right)\left(\frac{2}{x} + 3\right) $$

**Using the Difference of Squares Formula:**
$$
\left(\frac{2}{x} - 3\right)\left(\frac{2}{x} + 3\right) = \left(\frac{2}{x}\right)^{2} - 3^{2} = \frac{4}{x^{2}} - 9
$$

**Simplified Form:**
$$
\frac{4}{x^{2}} - 9
$$

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**5.** $$ \left(\frac{1}{3}x - \frac{3}{4}y\right)^{2} $$

**Expansion:**
$$
\left(\frac{1}{3}x - \frac{3}{4}y\right)^{2} = \left(\frac{1}{3}x\right)^{2} - 2 \cdot \frac{1}{3}x \cdot \frac{3}{4}y + \left(\frac{3}{4}y\right)^{2}
$$
$$
= \frac{1}{9}x^{2} - \frac{6}{12}xy + \frac{9}{16}y^{2}
$$
$$
= \frac{1}{9}x^{2} - \frac{1}{2}xy + \frac{9}{16}y^{2}
$$

**Simplified Form:**
$$
\frac{1}{9}x^{2} - \frac{1}{2}xy + \frac{9}{16}y^{2}
$$

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If you need further assistance or have specific instructions for these problems, feel free to let me know!
Here are the expanded forms of the given expressions: 1. $$ \left(a + \frac{1}{a}\right)^{2} = a^{2} + 2 + \frac{1}{a^{2}} $$ 2. $$ \left(2a + \frac{1}{2}\right)^{2} = 4a^{2} + 2a + \frac{1}{4} $$ 3. $$ \left(\frac{1}{x} - 3\right)\left(\frac{3}{x} + 1\right) = \frac{3}{x^{2}} - \frac{8}{x} - 3 $$ 4. $$ \left(\frac{2}{x} - 3\right)\left(\frac{2}{x} + 3\right) = \frac{4}{x^{2}} - 9 $$ 5. $$ \left(\frac{1}{3}x - \frac{3}{4}y\right)^{2} = \frac{1}{9}x^{2} - \frac{1}{2}xy + \frac{9}{16}y^{2} $$

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