Despejar variables - Ecuaciones. Tener en cuenta: las reglas o principios. I. Despejar \( x \) ó y en cada caso teniendo en cuenta la ecuación. 1. \( 2_{2} a(2 x-1)=3 \frac{9}{2} a \) 2. \( \frac{36+x}{x}=\frac{4 x}{36+x} \quad \begin{array}{c}\text { sar regla } \\ \text { general } \\ x=-b t i\end{array} \) 3. \( V_{1}^{2} x^{2}+V_{2} x^{2}=d^{2} \) general \( X=\frac{-b \pm}{b^{2}-4 a c} \) \( 2 a \) 4. \( 5^{2}=x^{2}+4^{2} \) 5. \( \frac{30}{26}=\frac{2 x-1}{2 x-1} \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{36+x}{x}=\frac{4x}{36+x}\) - step1: Find the domain: \(\frac{36+x}{x}=\frac{4x}{36+x},x \in \left(-\infty,-36\right)\cup \left(-36,0\right)\cup \left(0,+\infty\right)\) - step2: Cross multiply: \(\left(36+x\right)\left(36+x\right)=x\times 4x\) - step3: Simplify the equation: \(\left(36+x\right)^{2}=4x^{2}\) - step4: Expand the expression: \(1296+72x+x^{2}=4x^{2}\) - step5: Move the expression to the left side: \(1296+72x+x^{2}-4x^{2}=0\) - step6: Subtract the terms: \(1296+72x-3x^{2}=0\) - step7: Factor the expression: \(3\left(36-x\right)\left(12+x\right)=0\) - step8: Divide the terms: \(\left(36-x\right)\left(12+x\right)=0\) - step9: Separate into possible cases: \(\begin{align}&36-x=0\\&12+x=0\end{align}\) - step10: Solve the equation: \(\begin{align}&x=36\\&x=-12\end{align}\) - step11: Check if the solution is in the defined range: \(\begin{align}&x=36\\&x=-12\end{align},x \in \left(-\infty,-36\right)\cup \left(-36,0\right)\cup \left(0,+\infty\right)\) - step12: Find the intersection: \(\begin{align}&x=36\\&x=-12\end{align}\) - step13: Rewrite: \(x_{1}=-12,x_{2}=36\) Solve the equation \( \frac{30}{26}=\frac{2 x-1}{2 x-1} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{30}{26}=\frac{2x-1}{2x-1}\) - step1: Reduce the fraction: \(\frac{15}{13}=1\) - step2: Multiply both sides of the equation by LCD: \(\frac{15}{13}\times 13=1\times 13\) - step3: Simplify the equation: \(15=13\) - step4: The statement is false: \(x \in \varnothing \) Solve the equation \( 5^{2}=x^{2}+4^{2} \). Solve the quadratic equation by following steps: - step0: Solve using square roots: \(5^{2}=x^{2}+4^{2}\) - step1: Evaluate the power: \(5^{2}=x^{2}+16\) - step2: Swap the sides: \(x^{2}+16=5^{2}\) - step3: Add or subtract both sides: \(x^{2}=9\) - step4: Simplify the expression: \(x=\pm \sqrt{9}\) - step5: Simplify the expression: \(x=\pm 3\) - step6: Separate into possible cases: \(\begin{align}&x=3\\&x=-3\end{align}\) - step7: Rewrite: \(x_{1}=-3,x_{2}=3\) Solve the equation \( V_{1}^{2} x^{2}+V_{2} x^{2}=d^{2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(V_{1}^{2}x^{2}+V_{2}x^{2}=d^{2}\) - step1: Simplify: \(\left(V_{1}^{2}+V_{2}\right)x^{2}=d^{2}\) - step2: Divide both sides: \(\frac{\left(V_{1}^{2}+V_{2}\right)x^{2}}{V_{1}^{2}+V_{2}}=\frac{d^{2}}{V_{1}^{2}+V_{2}}\) - step3: Divide the numbers: \(x^{2}=\frac{d^{2}}{V_{1}^{2}+V_{2}}\) - step4: Simplify the expression: \(x=\pm \sqrt{\frac{d^{2}}{V_{1}^{2}+V_{2}}}\) - step5: Simplify: \(x=\pm \frac{d\sqrt{V_{1}^{2}+V_{2}}}{V_{1}^{2}+V_{2}}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{d\sqrt{V_{1}^{2}+V_{2}}}{V_{1}^{2}+V_{2}}\\&x=-\frac{d\sqrt{V_{1}^{2}+V_{2}}}{V_{1}^{2}+V_{2}}\end{align}\) Vamos a despejar las variables en cada una de las ecuaciones dadas, siguiendo las reglas y principios de álgebra. ### 1. Ecuación: \( 2_{2} a(2 x-1)=3 \frac{9}{2} a \) Parece que hay un error en la notación de la ecuación. La notación \( 2_{2} a \) no es clara. Por favor, verifica la ecuación y proporciona la forma correcta para que pueda resolverla. ### 2. Ecuación: \( \frac{36+x}{x}=\frac{4 x}{36+x} \) Al resolver esta ecuación, encontramos que: \[ x_{1} = -12, \quad x_{2} = 36 \] ### 3. Ecuación: \( V_{1}^{2} x^{2}+V_{2} x^{2}=d^{2} \) Despejando \( x \), obtenemos: \[ x = \frac{d \sqrt{V_{1}^{2}+V_{2}}}{V_{1}^{2}+V_{2}} \quad \text{y} \quad x = -\frac{d \sqrt{V_{1}^{2}+V_{2}}}{V_{1}^{2}+V_{2}} \] ### 4. Ecuación: \( 5^{2}=x^{2}+4^{2} \) Al resolver esta ecuación, encontramos que: \[ x_{1} = -3, \quad x_{2} = 3 \] ### 5. Ecuación: \( \frac{30}{26}=\frac{2 x-1}{2 x-1} \) Al resolver esta ecuación, encontramos que: \[ x \in \varnothing \] Esto significa que no hay solución para esta ecuación, ya que el denominador se anula. Si necesitas más aclaraciones o si hay alguna ecuación que quieras revisar, házmelo saber.