Question

01. Să se rezolve ecuațiile:
b.)​

Answer 1

 

$\displaystyle\bf\\01)~Sa~se~rezolve~ecuatiile:\\b)\\\\9^{2x+1}-3^{x+1}-6=0\\\\\Big(3^2\Big)^{2x+1}-3^{x+1}-6=0\\\\3^{2(2x+1)}-3^{x+1}=6\\\\3^{4x+2}-3^{x+1}=6\\\\Avem~o~diferenta~de~puteri~pentru~care~nu~avem~formula.\\Termenul~liber~din~drepta~nu~este~o~putere~a~lui~3.\\\\Rezulta~ca~trebuie~sa~facem~un~artificiu:\\\\Pe~6~il~scriem~ca~o~diferenta~dintre~2~puteri~ale~lui~3.\\\\6=9-3=3^2-3^1\\\\3^{4x+2}-3^{x+1}=3^2-3^1$

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$\displaystyle\bf\\Descompunem~ecuatia~in~2~ecuatii:\\\\3^{4x+2}=3^2\\\\3^{x+1}=3^1\\\\\textbf{Rezolvam ecuatiile separat si ar trebui sa obtinem aceasi solutie.}\\\\Rezolvare~E1:\\\\3^{4x+2}=3^2\\4x+2=2\\4x=2-2\\4x=0\\\\x=\frac{0}{4} \\\\\boxed{\bf x=0}\\\\Rezolvare~E2:\\\\3^{x+1}=3^1\\x+1=1\\x=1-1\\\boxed{\bf x=0}\\\\Avem~aceeasi~solutie.\\\\Verificare:\\\\9^{2x+1}-3^{x+1}-6=0\\x=0\\9^{2\cdot0+1}-3^{0+1}-6=0\\9^{0+1}-3^{0+1}-6=0\\9^{1}-3^{1}-6=0\\9-3-6=0\\6-6=0\\Corect$