https://file.aiquickdraw.com/mathbot/file/17260258736958zejtikr.jpg

Let's go through the mathematical expressions from the image and solve them one by one.

### 1. Solve:

#### a. 
\[
\left( \frac{x^{-5}y^4}{xy^3} \right)^{-2} \cdot \left( \frac{x^7y^{-3}}{x^{-4}y^6} \right)^{-\frac{1}{2}}, \quad x \neq 0, y \neq 0
\]

**Step-by-step:**

1. Simplify the first term:
   \[
   \frac{x^{-5}y^4}{xy^3} = x^{-5-1} y^{4-3} = x^{-6} y^1 = \frac{y}{x^6}
   \]
   Therefore:
   \[
   \left( \frac{y}{x^6} \right)^{-2} = \frac{x^{12}}{y^2}
   \]

2. Simplify the second term:
   \[
   \frac{x^7y^{-3}}{x^{-4}y^6} = x^{7+4} y^{-3-6} = x^{11} y^{-9} = \frac{x^{11}}{y^9}
   \]
   Then:
   \[
   \left( \frac{x^{11}}{y^9} \right)^{-\frac{1}{2}} = \frac{y^{\frac{9}{2}}}{x^{\frac{11}{2}}}
   \]

3. Multiply the two results:
   \[
   \frac{x^{12}}{y^2} \cdot \frac{y^{\frac{9}{2}}}{x^{\frac{11}{2}}}
   \]
   Simplifying:
   \[
   x^{12 - \frac{11}{2}} \cdot y^{\frac{9}{2} - 2} = x^{\frac{24}{2} - \frac{11}{2}} y^{\frac{9}{2} - \frac{4}{2}} = x^{\frac{13}{2}} y^{\frac{5}{2}}
   \]
   Final result for **a**:
   \[
   x^{\frac{13}{2}} y^{\frac{5}{2}}
   \]

---

#### b.
\[
\frac{\left( m^{10}n^{-2} \right)^3 \cdot \left( m^5n^{-5} \right)^3}{mn}, \quad m \neq 0, n \neq 0
\]

**Step-by-step:**

1. Simplify the numerator:
   \[
   \left( m^{10}n^{-2} \right)^3 = m^{30}n^{-6}, \quad \left( m^5n^{-5} \right)^3 = m^{15}n^{-15}
   \]
   So the numerator becomes:
   \[
   m^{30}n^{-6} \cdot m^{15}n^{-15} = m^{45}n^{-21}
   \]

2. Simplify the fraction:
   \[
   \frac{m^{45}n^{-21}}{mn} = m^{45-1}n^{-21-1} = m^{44}n^{-22}
   \]
   Final result for **b**:
   \[
   \frac{m^{44}}{n^{22}}
   \]

---

#### c.
\[
\frac{p + q}{\sqrt{p} - \sqrt{q}}, \quad p \neq 0, q \neq 0
\]

**Step-by-step:**

1. Multiply both the numerator and the denominator by the conjugate of the denominator:
   \[
   \frac{p + q}{\sqrt{p} - \sqrt{q}} \cdot \frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} + \sqrt{q}} = \frac{(p+q)(\sqrt{p} + \sqrt{q})}{(\sqrt{p})^2 - (\sqrt{q})^2}
   \]
   Simplify the denominator:
   \[
   (\sqrt{p})^2 - (\sqrt{q})^2 = p - q
   \]
   So we have:
   \[
   \frac{(p + q)(\sqrt{p} + \sqrt{q})}{p - q}
   \]
   Final result for **c**:
   \[
   \frac{(p + q)(\sqrt{p} + \sqrt{q})}{p - q}
   \]

---

#### d.
\[
\log \left( \frac{t + 6}{36 - t^2} \right)
\]

**Step-by-step:**

1. Notice that \(36 - t^2 = (6 - t)(6 + t)\), so we can rewrite the expression as:
   \[
   \log \left( \frac{t + 6}{(6 + t)(6 - t)} \right)
   \]
   Simplify the fraction:
   \[
   \frac{t + 6}{(6 + t)(6 - t)} = \frac{1}{6 - t}
   \]
   Therefore:
   \[
   \log \left( \frac{1}{6 - t} \right)
   \]
   Using the logarithmic property \( \log \left( \frac{1}{x} \right) = -\log(x) \), we get:
   \[
   -\log(6 - t)
   \]
   Final result for **d**:
   \[
   -\log(6 - t)
   \]

---

Let me know if you'd like further details on any of the steps or have any questions! Here are some related problems you might find helpful:

1. How would you simplify an expression with negative exponents?
2. What are the properties of logarithms that help in simplifying complex expressions?
3. How does the conjugate method work when rationalizing denominators?
4. What are some other techniques to simplify fractions with exponents?
5. Can you identify other types of indeterminate forms in logarithmic expressions?

**Tip:** Always try to simplify each term step-by-step before combining them for an easier approach!

Selesaikanlah:

a. \(\left(\frac{x^{-5}y^4}{xy^3}\right)^{-2} \cdot \left(\frac{x^7y^{-3}}{x^{-4}y^6}\right)^{-\frac{1}{2}}\), \quad x \neq 0, y \neq 0\)

b. \(\frac{\left(m^{10}n^{-2}\right)^3 \cdot \left(m^5n^{-5}\right)^3}{mn}\), \quad m \neq 0, n \neq 0\)

c. \(\frac{p + q}{\sqrt{p} - \sqrt{q}}\), \quad p \neq 0, q \neq 0\)

d. \(\log\left(\frac{t + 6}{36 - t^2}\right)\)

Explore the step-by-step solutions for solving complex algebraic expressions involving exponents and logarithms. This problem set covers exponent rules, logarithmic identities, and the conjugate method for rationalizing denominators. Suitable for students in Grades 10-12.

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