36 Résoudre dans \( \mathbb{R} \times \mathbb{R} \) les systèmes d'équations suivants : (1) \( \left\{\begin{aligned} x+y & =-1 \\ x y & =-1\end{aligned}\right. \) (4) \( \left\{\begin{aligned} \frac{x^{2}}{y}+\frac{y^{2}}{x} & =0 \\ x y & =-1\end{aligned}\right. \) (2) \( \left\{\begin{aligned} x^{2}+y^{2} & =5 \\ x y & =-2\end{aligned}\right. \) (5) \( \left\{\begin{aligned} x^{4}+y^{4} & =17 \\ x y & =2\end{aligned}\right. \) (3) \( \left\{\begin{array}{l}x^{2}+y^{2}=20 \\ \frac{x}{y}+\frac{y}{x}=\frac{5}{2}\end{array}\right. \) (6) \( \left\{\begin{aligned} x-y & =2 \\ x y & =35\end{aligned}\right. \) 37 Résoudre dans \( \mathbb{R} \) les équations suivantes : (1) \( 2 x|x|-6 x-2=0 \); (2) \( 2 x^{2}-3 x+5=|x-1| \) 38 On considère l'équation :

Solve the equation by following steps: - step0: Solve for \(x\): \(2x^{2}-3x+5=\left|x-1\right|\) - step1: Move the expression to the left side: \(2x^{2}-3x+5-\left|x-1\right|=0\) - step2: Separate into possible cases: \(\begin{align}&2x^{2}-3x+5-\left(x-1\right)=0,x-1\geq 0\\&2x^{2}-3x+5-\left(-\left(x-1\right)\right)=0,x-1<0\end{align}\) - step3: Solve the equation: \(\begin{align}&x \notin \mathbb{R},x\geq 1\\&x \notin \mathbb{R},x<1\end{align}\) - step4: Find the intersection: \(\begin{align}&x \notin \mathbb{R}\\&x \notin \mathbb{R}\end{align}\) - step5: Find the union: \(x \notin \mathbb{R}\) Solve the system of equations \( x^{4}+y^{4}=17;xy=2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{4}+y^{4}=17\\xy=2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{4}+y^{4}=17\\x=\frac{2}{y}\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(\frac{2}{y}\right)^{4}+y^{4}=17\) - step3: Simplify: \(\frac{16+y^{8}}{y^{4}}=17\) - step4: Cross multiply: \(16+y^{8}=y^{4}\times 17\) - step5: Simplify the equation: \(16+y^{8}=17y^{4}\) - step6: Move the expression to the left side: \(16+y^{8}-17y^{4}=0\) - step7: Factor the expression: \(\left(2-y\right)\left(1-y\right)\left(1+y\right)\left(2+y\right)\left(1+y^{2}\right)\left(4+y^{2}\right)=0\) - step8: Separate into possible cases: \(2-y=0\cup 1-y=0\cup 1+y=0\cup 2+y=0\cup 1+y^{2}=0\cup 4+y^{2}=0\) - step9: Solve the equation: \(y=2\cup y=1\cup y=-1\cup y=-2\cup y \notin \mathbb{R}\cup y \notin \mathbb{R}\) - step10: Find the union: \(y=2\cup y=1\cup y=-1\cup y=-2\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{2}{y}\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{y}\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{y}\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{y}\\y=-2\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=-2\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-1,-2\right)\cup \left(x,y\right) = \left(-2,-1\right)\cup \left(x,y\right) = \left(1,2\right)\cup \left(x,y\right) = \left(2,1\right)\) Solve the system of equations \( x-y=2;xy=35 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-y=2\\xy=35\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=2+y\\xy=35\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(2+y\right)y=35\) - step3: Simplify: \(y\left(2+y\right)=35\) - step4: Expand the expression: \(2y+y^{2}=35\) - step5: Move the expression to the left side: \(2y+y^{2}-35=0\) - step6: Factor the expression: \(\left(y-5\right)\left(y+7\right)=0\) - step7: Separate into possible cases: \(\begin{align}&y-5=0\\&y+7=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=5\\&y=-7\end{align}\) - step9: Calculate: \(y=5\cup y=-7\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=2+y\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=2+y\\y=-7\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=7\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=-5\\y=-7\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-5\\y=-7\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=5\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-5\\y=-7\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=5\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-5,-7\right)\cup \left(x,y\right) = \left(7,5\right)\) Solve the equation \( 2x|x|-6x-2=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2x\left|x\right|-6x-2=0\) - step1: Separate into possible cases: \(\begin{align}&2x\times x-6x-2=0,x\geq 0\\&2x\left(-x\right)-6x-2=0,x<0\end{align}\) - step2: Solve the equation: \(\begin{align}&\begin{align}&x=\frac{3+\sqrt{13}}{2}\\&x=\frac{3-\sqrt{13}}{2}\end{align},x\geq 0\\&\begin{align}&x=\frac{-3+\sqrt{5}}{2}\\&x=-\frac{3+\sqrt{5}}{2}\end{align},x<0\end{align}\) - step3: Find the intersection: \(\begin{align}&x=\frac{3+\sqrt{13}}{2}\\&x=\frac{-3+\sqrt{5}}{2}\\&x=-\frac{3+\sqrt{5}}{2}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{3+\sqrt{5}}{2},x_{2}=\frac{-3+\sqrt{5}}{2},x_{3}=\frac{3+\sqrt{13}}{2}\) Solve the system of equations \( x^{2}+y^{2}=20;\frac{x}{y}+\frac{y}{x}=\frac{5}{2} \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}+y^{2}=20\\\frac{x}{y}+\frac{y}{x}=\frac{5}{2}\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{2}+y^{2}=20\\x=\frac{y}{2}\cup x=2y\end{array}\right.\) - step2: Evaluate: \(\left\{ \begin{array}{l}x^{2}+y^{2}=20\\x=\frac{y}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x^{2}+y^{2}=20\\x=2y\end{array}\right.\) - step3: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=-4\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=-4\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=2\end{array}\right.\) - step4: Rearrange the terms: \(\left\{ \begin{array}{l}x=-2\\y=-4\end{array}\right.\cup \left\{ \begin{array}{l}x=-4\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=2\end{array}\right.\) - step5: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=-4\end{array}\right.\cup \left\{ \begin{array}{l}x=-4\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=2\end{array}\right.\) - step6: Rewrite: \(\left(x,y\right) = \left(-2,-4\right)\cup \left(x,y\right) = \left(-4,-2\right)\cup \left(x,y\right) = \left(2,4\right)\cup \left(x,y\right) = \left(4,2\right)\) Solve the system of equations \( \frac{x^{2}}{y}+\frac{y^{2}}{x}=0;xy=-1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{x^{2}}{y}+\frac{y^{2}}{x}=0\\xy=-1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}\frac{x^{2}}{y}+\frac{y^{2}}{x}=0\\x=-\frac{1}{y}\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\frac{\left(-\frac{1}{y}\right)^{2}}{y}+\frac{y^{2}}{-\frac{1}{y}}=0\) - step3: Simplify: \(\frac{1}{y^{3}}-y^{3}=0\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{1}{y^{3}}-y^{3}\right)y^{3}=0\times y^{3}\) - step5: Simplify the equation: \(1-y^{6}=0\) - step6: Rewrite the expression: \(-y^{6}=-1\) - step7: Change the signs: \(y^{6}=1\) - step8: Simplify the expression: \(y=\pm \sqrt[6]{1}\) - step9: Simplify: \(y=\pm 1\) - step10: Separate into possible cases: \(y=1\cup y=-1\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=-\frac{1}{y}\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{1}{y}\\y=-1\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-1\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-1\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-1,1\right)\cup \left(x,y\right) = \left(1,-1\right)\) Solve the system of equations \( x+y=-1;xy=-1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=-1\\xy=-1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-1-y\\xy=-1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(-1-y\right)y=-1\) - step3: Simplify: \(y\left(-1-y\right)=-1\) - step4: Expand the expression: \(-y-y^{2}=-1\) - step5: Move the expression to the left side: \(-y-y^{2}-\left(-1\right)=0\) - step6: Remove the parentheses: \(-y-y^{2}+1=0\) - step7: Rewrite in standard form: \(-y^{2}-y+1=0\) - step8: Multiply both sides: \(y^{2}+y-1=0\) - step9: Solve using the quadratic formula: \(y=\frac{-1\pm \sqrt{1^{2}-4\left(-1\right)}}{2}\) - step10: Simplify the expression: \(y=\frac{-1\pm \sqrt{5}}{2}\) - step11: Separate into possible cases: \(\begin{align}&y=\frac{-1+\sqrt{5}}{2}\\&y=\frac{-1-\sqrt{5}}{2}\end{align}\) - step12: Rewrite the fraction: \(\begin{align}&y=\frac{-1+\sqrt{5}}{2}\\&y=-\frac{1+\sqrt{5}}{2}\end{align}\) - step13: Evaluate the logic: \(y=\frac{-1+\sqrt{5}}{2}\cup y=-\frac{1+\sqrt{5}}{2}\) - step14: Rearrange the terms: \(\left\{ \begin{array}{l}x=-1-y\\y=\frac{-1+\sqrt{5}}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-1-y\\y=-\frac{1+\sqrt{5}}{2}\end{array}\right.\) - step15: Calculate: \(\left\{ \begin{array}{l}x=-\frac{1+\sqrt{5}}{2}\\y=\frac{-1+\sqrt{5}}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{-1+\sqrt{5}}{2}\\y=-\frac{1+\sqrt{5}}{2}\end{array}\right.\) - step16: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{1+\sqrt{5}}{2}\\y=\frac{-1+\sqrt{5}}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{-1+\sqrt{5}}{2}\\y=-\frac{1+\sqrt{5}}{2}\end{array}\right.\) - step17: Rewrite: \(\left(x,y\right) = \left(-\frac{1+\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}\right)\cup \left(x,y\right) = \left(\frac{-1+\sqrt{5}}{2},-\frac{1+\sqrt{5}}{2}\right)\) Solve the system of equations \( x^{2}+y^{2}=5;xy=-2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}+y^{2}=5\\xy=-2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{2}+y^{2}=5\\x=-\frac{2}{y}\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(-\frac{2}{y}\right)^{2}+y^{2}=5\) - step3: Simplify: \(\frac{4+y^{4}}{y^{2}}=5\) - step4: Cross multiply: \(4+y^{4}=y^{2}\times 5\) - step5: Simplify the equation: \(4+y^{4}=5y^{2}\) - step6: Move the expression to the left side: \(4+y^{4}-5y^{2}=0\) - step7: Factor the expression: \(\left(2-y\right)\left(1-y\right)\left(1+y\right)\left(2+y\right)=0\) - step8: Separate into possible cases: \(2-y=0\cup 1-y=0\cup 1+y=0\cup 2+y=0\) - step9: Solve the equation: \(y=2\cup y=1\cup y=-1\cup y=-2\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=-\frac{2}{y}\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2}{y}\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2}{y}\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2}{y}\\y=-2\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-2\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-1\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-1\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-1,2\right)\cup \left(x,y\right) = \left(-2,1\right)\cup \left(x,y\right) = \left(1,-2\right)\cup \left(x,y\right) = \left(2,-1\right)\) Voici les solutions pour les systèmes d'équations et les équations demandées : ### Systèmes d'équations 1. Pour le système \( \left\{\begin{aligned} x+y & =-1 \\ x y & =-1\end{aligned}\right. \): - Solutions : \[ (x,y) = \left(-\frac{1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right) \cup (x,y) = \left(\frac{-1+\sqrt{5}}{2}, -\frac{1+\sqrt{5}}{2}\right) \] 2. Pour le système \( \left\{\begin{aligned} \frac{x^{2}}{y}+\frac{y^{2}}{x} & =0 \\ x y & =-1\end{aligned}\right. \): - Solutions : \[ (x,y) = (-1,1) \cup (x,y) = (1,-1) \] 3. Pour le système \( \left\{\begin{aligned} x^{2}+y^{2} & =5 \\ x y & =-2\end{aligned}\right. \): - Solutions : \[ (x,y) = (-1,2) \cup (x,y) = (-2,1) \cup (x,y) = (1,-2) \cup (x,y) = (2,-1) \] 4. Pour le système \( \left\{\begin{aligned} x^{4}+y^{4} & =17 \\ x y & =2\end{aligned}\right. \): - Solutions : \[ (x,y) = (-1,-2) \cup (x,y) = (-2,-1) \cup (x,y) = (1,2) \cup (x,y) = (2,1) \] 5. Pour le système \( \left\{\begin{aligned} x^{2}+y^{2} & =20 \\ \frac{x}{y}+\frac{y}{x}=\frac{5}{2}\end{aligned}\right. \): - Solutions : \[ (x,y) = (-2,-4) \cup (x,y) = (-4,-2) \cup (x,y) = (2,4) \cup (x,y) = (4,2) \] 6. Pour le système \( \left\{\begin{aligned} x-y & =2 \\ x y & =35\end{aligned}\right. \): - Solutions : \[ (x,y) = (-5,-7) \cup (x,y) = (7,5) \] ### Équations 1. Pour l'équation \( 2 x|x|-6 x-2=0 \): - Solutions : \[ x_{1}=-\frac{3+\sqrt{5}}{2}, \quad x_{2}=\frac{-3+\sqrt{5}}{2}, \quad x_{3}=\frac{3+\sqrt{13}}{2} \] 2. Pour l'équation \( 2 x^{2}-3 x+5=|x-1| \): - Il n'y a pas de solution réelle : \[ x \notin \mathbb{R \text{ (pas de solution réelle)}} \] Si vous avez besoin de plus d'informations ou d'autres questions, n'hésitez pas à demander !