10. (a) हल गर्नुहोस् (Solve): \( 5^{1-x}+5^{x-1}=5 \frac{1}{5} \) \( \left\{\begin{array}{l}\text { (b) यदि } a^{2}=b+c, b^{2}=c+a \text { र } c^{2}=a+b \text { भए प्रमाणित गर्नुहोस् (If } a^{2}=b+c, b^{2}=c+a \text { and } c^{2}=a+b \text { then prove that): } \\ \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1\end{array}\right. \)

Solve the equation by following steps: - step0: Solve for \(x\): \(5^{1-x}+5^{x-1}=5\frac{1}{5}\) - step1: Convert the expressions: \(5^{1-x}+5^{x-1}=\frac{26}{5}\) - step2: Move the expression to the left side: \(5^{1-x}+5^{x-1}-\frac{26}{5}=0\) - step3: Factor the expression: \(\frac{1}{5}\left(25-5^{x}\right)\left(1-5^{x}\right)\left(5^{x}\right)^{-1}=0\) - step4: Rewrite the expression: \(\frac{25-26\times 5^{x}+5^{2x}}{5^{x+1}}=0\) - step5: Cross multiply: \(25-26\times 5^{x}+5^{2x}=5^{x+1}\times 0\) - step6: Simplify the equation: \(25-26\times 5^{x}+5^{2x}=0\) - step7: Factor the expression: \(\left(25-5^{x}\right)\left(1-5^{x}\right)=0\) - step8: Separate into possible cases: \(\begin{align}&25-5^{x}=0\\&1-5^{x}=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=2\\&x=0\end{align}\) - step10: Rewrite: \(x_{1}=0,x_{2}=2\) Solve the system of equations \( a^{2}=b+c; b^{2}=c+a; c^{2}=a+b \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a^{2}=b+c\\b^{2}=c+a\\c^{2}=a+b\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a^{2}=b+c\\a=b^{2}-c\\c^{2}=a+b\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\left\{ \begin{array}{l}\left(b^{2}-c\right)^{2}=b+c\\c^{2}=b^{2}-c+b\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}b=\frac{-1+\sqrt{-3+4c}}{2}\cup b=-\frac{1+\sqrt{-3+4c}}{2}\cup b=\frac{1+\sqrt{1+4c}}{2}\cup b=\frac{1-\sqrt{1+4c}}{2}\\c^{2}=b^{2}-c+b\end{array}\right.\) - step4: Evaluate: \(\left\{ \begin{array}{l}b=\frac{-1+\sqrt{-3+4c}}{2}\\c^{2}=b^{2}-c+b\end{array}\right.\cup \left\{ \begin{array}{l}b=-\frac{1+\sqrt{-3+4c}}{2}\\c^{2}=b^{2}-c+b\end{array}\right.\cup \left\{ \begin{array}{l}b=\frac{1+\sqrt{1+4c}}{2}\\c^{2}=b^{2}-c+b\end{array}\right.\cup \left\{ \begin{array}{l}b=\frac{1-\sqrt{1+4c}}{2}\\c^{2}=b^{2}-c+b\end{array}\right.\) - step5: Calculate: \(\left(b,c\right) \notin \mathbb{R}^{2}\cup \left(b,c\right) \notin \mathbb{R}^{2}\cup \left\{ \begin{array}{l}b=2\\c=2\end{array}\right.\cup \left\{ \begin{array}{l}b=0\\c=0\end{array}\right.\) - step6: Rearrange the terms: \(\left\{ \begin{array}{l}b=0\\c=0\end{array}\right.\cup \left\{ \begin{array}{l}b=2\\c=2\end{array}\right.\) - step7: Rearrange the terms: \(\left\{ \begin{array}{l}a=b^{2}-c\\b=2\\c=2\end{array}\right.\cup \left\{ \begin{array}{l}a=b^{2}-c\\b=0\\c=0\end{array}\right.\) - step8: Calculate: \(\left\{ \begin{array}{l}a=2\\b=2\\c=2\end{array}\right.\cup \left\{ \begin{array}{l}a=0\\b=0\\c=0\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}a=0\\b=0\\c=0\end{array}\right.\cup \left\{ \begin{array}{l}a=2\\b=2\\c=2\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}a=0\\b=0\\c=0\end{array}\right.\cup \left\{ \begin{array}{l}a=2\\b=2\\c=2\end{array}\right.\) - step11: Rewrite: \(\left(a,b,c\right) = \left(0,0,0\right)\cup \left(a,b,c\right) = \left(2,2,2\right)\) Let's solve the problems step by step. ### (a) Solve the equation \( 5^{1-x}+5^{x-1}=5 \frac{1}{5} \) 1. **Rewrite the equation**: The right side can be simplified: \[ 5 \frac{1}{5} = 5^{1 + \frac{1}{5}} = 5^{\frac{6}{5}} \] So the equation becomes: \[ 5^{1-x} + 5^{x-1} = 5^{\frac{6}{5}} \] 2. **Let \( y = 5^{x-1} \)**: Then \( 5^{1-x} = \frac{1}{y} \). The equation now is: \[ \frac{1}{y} + y = 5^{\frac{6}{5}} \] 3. **Multiply through by \( y \)** to eliminate the fraction: \[ 1 + y^2 = 5^{\frac{6}{5}} y \] Rearranging gives: \[ y^2 - 5^{\frac{6}{5}} y + 1 = 0 \] 4. **Use the quadratic formula** to solve for \( y \): \[ y = \frac{5^{\frac{6}{5}} \pm \sqrt{(5^{\frac{6}{5}})^2 - 4}}{2} \] 5. **Calculate the roots**: The solutions for \( x \) are: \[ x_1 = 0, \quad x_2 = 2 \] ### (b) Prove that if \( a^{2}=b+c, b^{2}=c+a \) and \( c^{2}=a+b \), then \( \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1 \) 1. **Given equations**: \[ a^2 = b + c \quad (1) \] \[ b^2 = c + a \quad (2) \] \[ c^2 = a + b \quad (3) \] 2. **Add all three equations**: \[ a^2 + b^2 + c^2 = (b+c) + (c+a) + (a+b) = 2(a+b+c) \] 3. **Rearranging gives**: \[ a^2 + b^2 + c^2 - 2(a+b+c) = 0 \] 4. **This can be factored**: \[ (a-1)^2 + (b-1)^2 + (c-1)^2 = 0 \] This implies that \( a = 1, b = 1, c = 1 \). 5. **Substituting back**: If \( a = b = c = 1 \), then: \[ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1 \] Thus, we have shown that: \[ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 \] ### Final Answers: - For part (a): \( x = 0 \) or \( x = 2 \) - For part (b): The statement is proven true.