1rBTX 02. POLINOMIS/1.2.0 Repàs de polinomis 1.2.0.81Q SigneFracció ax+b/x+c Aleat_1 Resol: #e #v 0

Format de la resposta:

Entre dos nombres: «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mo»§lt;«/mo»«mi»x«/mi»«mo»§#8743;«/mo»«mi»x«/mi»«mo»§lt;«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»(«/mo»«mo»§#8743;«/mo»«mo»=«/mo»«mi»i«/mi»«mo»)«/mo»«/math»

Altra situació: «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»§lt;«/mo»«mo»-«/mo»«mn»2«/mn»«mo»§#8744;«/mo»«mi»x«/mi»«mo»§#62;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»(«/mo»«mo»§#8744;«/mo»«mo»=«/mo»«mi»o«/mi»«mo»)«/mo»«/math»

]]> Els intervals solució queden definits per #sol

]]> 1.0000000 0.5000000 0 0 #sol

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xmlns="http://www.w3.org/1998/Math/MathML"><mi>s_2</mi></math></input><output><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo></math></output></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>sig</mi></math></input><output><math xmlns="http://www.w3.org/1998/Math/MathML"><ms>positiva</ms></math></output></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>sol</mi></math></input><output><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>></mo><mn>2</mn><mo>∧</mo><mi>x</mi><mo><</mo><mn>3</mn></math></output></command></group><group><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"/></input></command></group></session>]]></wirisCasSession><correctAnswers><correctAnswer>#sol</correctAnswer></correctAnswers><assertions><assertion name="equivalent_equations"/><assertion name="syntax_expression"/></assertions><options><option name="tolerance">10^(-4)</option><option name="relative_tolerance">false</option><option name="precision">4</option><option name="implicit_times_operator">false</option><option name="times_operator">·</option><option name="imaginary_unit">i</option></options><localData><data name="inputField">popupEditor</data><data name="gradeCompound">and</data><data name="gradeCompoundDistribution"></data><data name="casSession"/></localData></question> Cal fer la taula de signes per veure que:

pels valors de x inferiors a #x_1, l'expressió és #m_1 (#n_1 #a)
pels valors de x entre #x_1 i #x_2, l'expressió és #m_2 (#n_2 #a)
pels valors de x superiors a #x_2, l'expressió és #m_1 (#n_1 #a)

x	-∞	#x_1		#x_2	+∞
#n	#d3	0	#d4		#d5
#d	#d6		#d7	0	#d8
#e	#s_1	0	#s_2		#s_1

i l'enunciat demana que l'expressió sigui #sig

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