(1) $$ x-3 x^{\frac{1}{2}}+2=0 $$ (3) $$ x^{\frac{1}{2}}+3 x^{\frac{1}{4}}-10=0 $$

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Sure, let's solve each of the given equations step by step.

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### **Problem (1):** 
$$ x - 3x^{\frac{1}{2}} + 2 = 0 $$

#### **Solution:**

1. **Substitution:**
   Let $$ y = \sqrt{x} $$. Then, $$ x = y^2 $$.

2. **Rewrite the Equation:**
   Substitute $$ x = y^2 $$ into the original equation:
   $$
   y^2 - 3y + 2 = 0
   $$

3. **Solve the Quadratic Equation:**
   $$
   y^2 - 3y + 2 = 0
   $$
   Factor the quadratic:
   $$
   (y - 1)(y - 2) = 0
   $$
   So, $$ y = 1 $$ or $$ y = 2 $$.

4. **Back-Substitute to Find $$ x $$:**
   - If $$ y = 1 $$:
     $$
     \sqrt{x} = 1 \implies x = 1^2 = 1
     $$
   - If $$ y = 2 $$:
     $$
     \sqrt{x} = 2 \implies x = 2^2 = 4
     $$

5. **Solutions:**
   $$
   x = 1 \quad \text{and} \quad x = 4
   $$

---

### **Problem (3):** 
$$ x^{\frac{1}{2}} + 3x^{\frac{1}{4}} - 10 = 0 $$

#### **Solution:**

1. **Substitution:**
   Let $$ y = x^{\frac{1}{4}} $$. Then, $$ x^{\frac{1}{2}} = y^2 $$.

2. **Rewrite the Equation:**
   Substitute $$ y = x^{\frac{1}{4}} $$ into the original equation:
   $$
   y^2 + 3y - 10 = 0
   $$

3. **Solve the Quadratic Equation:**
   $$
   y^2 + 3y - 10 = 0
   $$
   Use the quadratic formula:
   $$
   y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   Here, $$ a = 1 $$, $$ b = 3 $$, and $$ c = -10 $$:
   $$
   y = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}
   $$
   So,
   $$
   y = \frac{4}{2} = 2 \quad \text{or} \quad y = \frac{-10}{2} = -5
   $$
   
4. **Discard Negative Solution:**
   Since $$ y = x^{\frac{1}{4}} $$ represents a real number (assuming $$ x $$ is real and non-negative), we discard $$ y = -5 $$.

5. **Back-Substitute to Find $$ x $$:**
   $$
   y = 2 \implies x^{\frac{1}{4}} = 2 \implies x = 2^4 = 16
   $$

6. **Solution:**
   $$
   x = 16
   $$

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### **Final Answers:**

1. **For $$ x - 3x^{\frac{1}{2}} + 2 = 0 $$:**
   $$
   x = 1 \quad \text{and} \quad x = 4
   $$

2. **For $$ x^{\frac{1}{2}} + 3x^{\frac{1}{4}} - 10 = 0 $$:**
   $$
   x = 16
   $$
The solutions are: 1. $$ x = 1 $$ and $$ x = 4 $$ 2. $$ x = 16 $$

(1) \( x-3 x^{\frac{1}{2}}+2=0 \) (3) \( x^{\frac{1}{2}}+3 x^{\frac{1}{4}}-10=0 \)

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