Her sien 1.3 Los op vir \( x \) 1.3 .1 x \( (x-4)=- \) 1.3 .2 5x+8<3x 1.4 Los gelyktydig \( z^{2+y} \)

Solve the equation by following steps: - step0: Solve for \(z\): \(z^{2+y}=0\) - step1: Calculate: \(z=0^{\frac{1}{2+y}}\) - step2: Simplify: \(z=0\) - step3: Check if the solution is in the defined range: \(z=0,z>0\) - step4: Find the intersection: \(z \in \varnothing \) Solve the equation \( 5x+8<3x \). Solve the inequality by following steps: - step0: Solve for \(x\): \(5x+8<3x\) - step1: Move the variable to the left side: \(5x+8-3x<0\) - step2: Subtract the terms: \(2x+8<0\) - step3: Move the constant to the right side: \(2x<0-8\) - step4: Remove 0: \(2x<-8\) - step5: Divide both sides: \(\frac{2x}{2}<\frac{-8}{2}\) - step6: Divide the numbers: \(x<-4\) Solve the equation \( x(x-4)=-1 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(x\left(x-4\right)=-1\) - step1: Expand the expression: \(x^{2}-4x=-1\) - step2: Move the expression to the left side: \(x^{2}-4x+1=0\) - step3: Solve using the quadratic formula: \(x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4}}{2}\) - step4: Simplify the expression: \(x=\frac{4\pm \sqrt{12}}{2}\) - step5: Simplify the expression: \(x=\frac{4\pm 2\sqrt{3}}{2}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{4+2\sqrt{3}}{2}\\&x=\frac{4-2\sqrt{3}}{2}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=2+\sqrt{3}\\&x=\frac{4-2\sqrt{3}}{2}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=2+\sqrt{3}\\&x=2-\sqrt{3}\end{align}\) - step9: Rewrite: \(x_{1}=2-\sqrt{3},x_{2}=2+\sqrt{3}\) Here are the solutions to the problems you provided: ### 1.3.1 Solve for \( x \) in the equation \( x(x-4) = -1 \) To solve the equation \( x(x-4) = -1 \), we can rearrange it to: \[ x^2 - 4x + 1 = 0 \] Using the quadratic formula, we find the solutions: \[ x_{1} = 2 - \sqrt{3}, \quad x_{2} = 2 + \sqrt{3} \] ### 1.3.2 Solve the inequality \( 5x + 8 < 3x \) Rearranging the inequality gives: \[ 5x - 3x < -8 \] This simplifies to: \[ 2x < -8 \] Dividing both sides by 2, we find: \[ x < -4 \] ### 1.4 Solve the equation \( z^{2+y} = 0 \) The equation \( z^{2+y} = 0 \) has no real solutions because a power of a real number is zero only when the base itself is zero. Therefore, we conclude: \[ z \in \varnothing \] ### Summary of Results: - For \( x(x-4) = -1 \): \( x_{1} = 2 - \sqrt{3}, \quad x_{2} = 2 + \sqrt{3} \) - For \( 5x + 8 < 3x \): \( x < -4 \) - For \( z^{2+y} = 0 \): No real solutions (\( z \in \varnothing \))