Question

URGEEEEEEENT! AJUTATI-MA VA ROG. Exercitiul 10

Answer 1

a). 2√(x-1) + √(x+3) = 2
Notam √(x-1)=a si √(x+3)=b, rezulta: 2a + b = 2.
Daca a=0, avem 2√(x-1)=0, deci x-1=0, de unde x=1.
Daca b=2, avem √(x+3) = 2, deci x+3=4, de unde x=1.
Verificare: 2√(1-1) + √(1+3) = 2*0 + √4 = 0 + 2 = 2, deci 2=2 (A)

b). ^3√(12-x) + ^3√(14+x) = 2, unde ^3√ este radical de ordin 3.
Notam ^3√(12-x)=c si ^3√(14+x)=d, rezulta: c + d = 2.
Daca c=-1 si d=3, avem -1+3=2, rezulta:
^3√(12-x) = -1, de unde x=13.
^3√(14+x) = 3, de unde x=13.
Verificare: ^3√(12-13) + ^3√(14+13) = ^3√(-1) + ^3√27 = -1 + 3 = 2, deci 2=2 (A).

c). ^3√(x+24) + √(12-x) = 6, unde ^3√ este radical de ordin 3.
Notam ^3√(x+24)=e si √(12-x)=f, rezulta e+f = 6.
Daca e=3, avem ^3√(x+24)=3, deci x+24=27, rezulta x=27-24=3.
Daca f=3, avem √(12-x)=3, deci 12-x=9, rezulta x=12-9=3.
Verificare: ^3√(3+24) + √(12-3) = ^3√27 + √9 = 3 + 3 = 6, deci 6=6 (A).