fisadelucru._ecuatiicuparametrureal_7
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Transcript of fisadelucru._ecuatiicuparametrureal_7
FISA DE LUCRU
FISA DE LUCRU (ECUATII DE GRADUL I CU O NECUNOSCUTA )1)Rezolvati in R urmatoarele ecuatii:
a)3 2(x 1)=x +5 b)3(x 1)+ 4(x 2)= 5 c)2x + 1)2- - (2x -1)2=x -5
d)2+ x -
EMBED Equation.3 (3 x)
EMBED Equation.3 = 1 e) = f)x - = x+ 32)Rezolvati in Z :a)5x -2= 7x -4 b) - = c)(x +1)(x +3) =(x+2)(x +4)d) -2 =4+ e) 2 =2 f)(2x +3)2= (3x -1)(3x +1)
3)Rezolvati in R ecuatiile:
a) =1 b) =5 c) =2 d) =
4) Rezolvati si analizati toate cazurile posibile in functie de parametrul real m pentru ecuatiile:
a)(m +1)x - 2x +3 =0 b)(2m +1)x +1 = 3x m +2 c)x-1 - m(x -2) = m +1
d)(3m -1)x = 9m2 -1 e)(2m +5) x +25 4m2 = 0 f)(m2 +1) x = m2 +1
g)(m2 1)x = 3x + 2(m +2) h) + =3 i) - =35) Aflati cel mai mic numar natural a pentru care urmatoarele ecuatii au solutii in Z:
a)4x -2 =x+ a b)1 x (1 -3x) =3a +5 c) 3x -2 = 1 (2x a)
d)2x +1 (2 -3x) =3x (1 a) e)5x +3 - 2(3a -1) =4 (2x -5) -3 (2a +5)
6)Fie ecuatia : (2m +1) x 4m +3 =0
a) determinati mZ astfel incat ecuatia sa admita solutii in Z .b) Pentru ce valori ,numere intregi ale lui m , solutiile ecuatiei sunt numere intregi si x ?7) Sa se rezolve ecuatia : + = in cazurile : m =2 si m =.
8)Se da ecuatia : mx +2m =x -5
a) sa se afle m R astfel incat ecuatia sa aiba solutia 3.b)sa se afle m R astfel incat ecuatia data sa fie echivalenta cu ecuatia :
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