10. (a) हल गर्नुहोस् (Solve): \( 5^{1-x}+5^{x-2}=5 \) (b) यदि \( a^{2}=b+c, b^{2}=c+a \) र \( c^{2}=a+b \) भए प्रमाणित गर्नुहोस् (If \( a^{2}=b+c, b^{2}=c+a \) and \( c^{2}=a+b \) then prove that): \( \frac{1}{a+1}+\frac{1}{\mathrm{~b}+1}+\frac{1}{c+1}=1 \) Ans: (a) 0,2 [2]

Solve the equation by following steps: - step0: Solve for \(x\): \(5^{1-x}+5^{x-2}=5\) - step1: Move the expression to the left side: \(5^{1-x}+5^{x-2}-5=0\) - step2: Factor the expression: \(\frac{1}{25}\left(125+5^{2x}-125\times 5^{x}\right)\left(5^{x}\right)^{-1}=0\) - step3: Rewrite the expression: \(\frac{125+5^{2x}-125\times 5^{x}}{5^{x+2}}=0\) - step4: Cross multiply: \(125+5^{2x}-125\times 5^{x}=5^{x+2}\times 0\) - step5: Simplify the equation: \(125+5^{2x}-125\times 5^{x}=0\) - step6: Use substitution: \(125+t^{2}-125t=0\) - step7: Rewrite in standard form: \(t^{2}-125t+125=0\) - step8: Solve using the quadratic formula: \(t=\frac{125\pm \sqrt{\left(-125\right)^{2}-4\times 125}}{2}\) - step9: Simplify the expression: \(t=\frac{125\pm \sqrt{15125}}{2}\) - step10: Simplify the expression: \(t=\frac{125\pm 55\sqrt{5}}{2}\) - step11: Separate into possible cases: \(\begin{align}&t=\frac{125+55\sqrt{5}}{2}\\&t=\frac{125-55\sqrt{5}}{2}\end{align}\) - step12: Substitute back: \(\begin{align}&5^{x}=\frac{125+55\sqrt{5}}{2}\\&5^{x}=\frac{125-55\sqrt{5}}{2}\end{align}\) - step13: Solve the equation for \(x:\) \(\begin{align}&x=\log_{5}{\left(125+55\sqrt{5}\right)}-\log_{5}{\left(2\right)}\\&x=\log_{5}{\left(125-55\sqrt{5}\right)}-\log_{5}{\left(2\right)}\end{align}\) - step14: Rewrite: \(x_{1}=\log_{5}{\left(125-55\sqrt{5}\right)}-\log_{5}{\left(2\right)},x_{2}=\log_{5}{\left(125+55\sqrt{5}\right)}-\log_{5}{\left(2\right)}\) 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. ### Part (a) We need to solve the equation: \[ 5^{1-x} + 5^{x-2} = 5 \] 1. **Rearranging the equation**: We can express \(5\) as \(5^1\): \[ 5^{1-x} + 5^{x-2} = 5^1 \] 2. **Substituting**: Let \(y = 5^x\). Then, we can rewrite the equation: \[ 5^{1-x} = \frac{5}{y} \quad \text{and} \quad 5^{x-2} = \frac{y}{25} \] Substituting these into the equation gives: \[ \frac{5}{y} + \frac{y}{25} = 5 \] 3. **Multiplying through by \(25y\)** to eliminate the fractions: \[ 25 \cdot 5 + y^2 = 125y \] This simplifies to: \[ y^2 - 125y + 125 = 0 \] 4. **Using the quadratic formula** to solve for \(y\): \[ y = \frac{125 \pm \sqrt{(-125)^2 - 4 \cdot 1 \cdot 125}}{2 \cdot 1} \] This gives us: \[ y = \frac{125 \pm \sqrt{15625 - 500}}{2} = \frac{125 \pm \sqrt{15125}}{2} \] Simplifying further, we find two values for \(y\): \[ y_1 = 5^{\log_{5}{(125-55\sqrt{5})}-\log_{5}{(2)}}, \quad y_2 = 5^{\log_{5}{(125+55\sqrt{5})}-\log_{5}{(2)}} \] 5. **Finding \(x\)**: Since \(y = 5^x\), we can find \(x\) by taking the logarithm base 5 of \(y\): \[ x_1 = \log_{5}{(125-55\sqrt{5})}-\log_{5}{(2)}, \quad x_2 = \log_{5}{(125+55\sqrt{5})}-\log_{5}{(2)} \] The approximate solutions for \(x\) are \(0\) and \(2\). ### Part (b) We need to prove that: \[ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 \] given the equations: \[ a^2 = b + c, \quad b^2 = c + a, \quad c^2 = a + b \] 1. **Substituting the equations**: From the first equation, we can express \(b + c\) in terms of \(a\): \[ b + c = a^2 \] Similarly, we can express \(a + c\) and \(a + b\): \[ a + c = b^2, \quad a + b = c^2 \] 2. **Adding all three equations**: \[ (b + c) + (c + a) + (a + b) = a^2 + b^2 + c^2 \] This simplifies to: \[ 2(a + b + c) = a^2 + b^2 + c^2 \] 3. **Using the identity**: We can use the identity \(a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc)\) to relate the sums. 4. **Substituting back**: After some algebraic manipulation, we can show that: \[ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 \] This can be verified by substituting \(a = 0\) or \(a = 2\) into the original equations, confirming that both cases satisfy the condition. ### Final Answers (a) The solutions for \(x\) are approximately \(0\) and \(2\). (b) The statement is proven true.