Solve for \( x \) : 1. \( x-5=\frac{6}{x} \) 2. \( \frac{x^{2}-3 x-7}{x^{2}-x-2}=\frac{x+2}{x+1}-1 \) 3. \( \frac{2}{x}-\frac{x+1}{x+2}=1 \)

Solve the equation \( x-5=\frac{6}{x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(x-5=\frac{6}{x}\) - step1: Find the domain: \(x-5=\frac{6}{x},x\neq 0\) - step2: Multiply both sides of the equation by LCD: \(\left(x-5\right)x=\frac{6}{x}\times x\) - step3: Simplify the equation: \(x^{2}-5x=6\) - step4: Move the expression to the left side: \(x^{2}-5x-6=0\) - step5: Factor the expression: \(\left(x-6\right)\left(x+1\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x-6=0\\&x+1=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=6\\&x=-1\end{align}\) - step8: Check if the solution is in the defined range: \(\begin{align}&x=6\\&x=-1\end{align},x\neq 0\) - step9: Find the intersection: \(\begin{align}&x=6\\&x=-1\end{align}\) - step10: Rewrite: \(x_{1}=-1,x_{2}=6\) Solve the equation \( \\frac{x^{2}-3x-7}{x^{2}-x-2}=\\frac{x+2}{x+1}-1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x^{2}-3x-7}{x^{2}-x-2}=\frac{x+2}{x+1}-1\) - step1: Find the domain: \(\frac{x^{2}-3x-7}{x^{2}-x-2}=\frac{x+2}{x+1}-1,x \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\frac{x^{2}-3x-7}{x^{2}-x-2}\times \left(x+1\right)\left(x-2\right)=\left(\frac{x+2}{x+1}-1\right)\left(x+1\right)\left(x-2\right)\) - step3: Simplify the equation: \(x^{2}-3x-7=x-2\) - step4: Move the expression to the left side: \(x^{2}-3x-7-\left(x-2\right)=0\) - step5: Subtract the terms: \(x^{2}-4x-5=0\) - step6: Factor the expression: \(\left(x-5\right)\left(x+1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x-5=0\\&x+1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=5\\&x=-1\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x=5\\&x=-1\end{align},x \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)\) - step10: Find the intersection: \(x=5\) Solve the equation \( \\frac{2}{x}-\\frac{x+1}{x+2}=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{2}{x}-\frac{x+1}{x+2}=1\) - step1: Find the domain: \(\frac{2}{x}-\frac{x+1}{x+2}=1,x \in \left(-\infty,-2\right)\cup \left(-2,0\right)\cup \left(0,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{2}{x}-\frac{x+1}{x+2}\right)x\left(x+2\right)=1\times x\left(x+2\right)\) - step3: Simplify the equation: \(x+4-x^{2}=x^{2}+2x\) - step4: Move the expression to the left side: \(x+4-x^{2}-\left(x^{2}+2x\right)=0\) - step5: Calculate: \(-x+4-2x^{2}=0\) - step6: Rewrite in standard form: \(-2x^{2}-x+4=0\) - step7: Multiply both sides: \(2x^{2}+x-4=0\) - step8: Solve using the quadratic formula: \(x=\frac{-1\pm \sqrt{1^{2}-4\times 2\left(-4\right)}}{2\times 2}\) - step9: Simplify the expression: \(x=\frac{-1\pm \sqrt{1^{2}-4\times 2\left(-4\right)}}{4}\) - step10: Simplify the expression: \(x=\frac{-1\pm \sqrt{33}}{4}\) - step11: Separate into possible cases: \(\begin{align}&x=\frac{-1+\sqrt{33}}{4}\\&x=\frac{-1-\sqrt{33}}{4}\end{align}\) - step12: Rewrite the fraction: \(\begin{align}&x=\frac{-1+\sqrt{33}}{4}\\&x=-\frac{1+\sqrt{33}}{4}\end{align}\) - step13: Check if the solution is in the defined range: \(\begin{align}&x=\frac{-1+\sqrt{33}}{4}\\&x=-\frac{1+\sqrt{33}}{4}\end{align},x \in \left(-\infty,-2\right)\cup \left(-2,0\right)\cup \left(0,+\infty\right)\) - step14: Find the intersection: \(\begin{align}&x=\frac{-1+\sqrt{33}}{4}\\&x=-\frac{1+\sqrt{33}}{4}\end{align}\) - step15: Rewrite: \(x_{1}=-\frac{1+\sqrt{33}}{4},x_{2}=\frac{-1+\sqrt{33}}{4}\) The solutions to the given equations are: 1. \( x-5=\frac{6}{x} \) has solutions \( x=-1 \) and \( x=6 \). 2. \( \frac{x^{2}-3x-7}{x^{2}-x-2}=\frac{x+2}{x+1}-1 \) has solution \( x=5 \). 3. \( \frac{2}{x}-\frac{x+1}{x+2}=1 \) has solutions \( x=-\frac{1+\sqrt{33}}{4} \) and \( x=\frac{-1+\sqrt{33}}{4} \).