Question

Bunaa
Ma poate ajuta si pe mine cu ceva la matematică, va rog!!
1) log5(3x+4)=2
2) Sa se calculeze 1/x1+1/x2,știind ca x1 si x2 sunt soluțiile ecuației x²-x-2=0

Answer 1

$1)\quad \log_{5}(3x+4) = 2\,\, \Leftrightarrow \,\, 3x+4 = 5^2 \,\, \Leftrightarrow\\ \\ \Leftrightarrow \,\, 3x+4 = 25 \,\,\Leftrightarrow\,\, 3x = 21\,\,\Leftrightarrow\,\, \boxed{x = 7}\in \Bigg[\text{D.V.A.} = \Big(-\dfrac{4}{3},\infty\Big)\Bigg]$

$\\ \\ 2)\quad x^2-x-2 = 0\Big|:x,\quad (x\neq 0) \,\,\Leftrightarrow \\ \\ \Leftrightarrow\,\, x - 1 - \dfrac{2}{x} = 0 \,\, \Leftrightarrow \dfrac{2}{x} = x-1\,\, \Leftrightarrow \,\, \dfrac{1}{x} = \dfrac{x-1}{2} \,\, \Leftrightarrow \\ \\ \Leftrightarrow\,\, \dfrac{1}{x_k} =\dfrac{x_k-1}{2}\,\, \Leftrightarrow \,\,\sum\limits_{k=1}^2\,\,\dfrac{1}{x_k} = \sum\limits_{k=1}^2\dfrac{x_k-1}{2}\,\,\Leftrightarrow$

$\Leftrightarrow \,\, \dfrac{1}{x_1}+\dfrac{1}{x_2} = \dfrac{x_1-1}{2}+\dfrac{x_2-1}{2}\,\, \Leftrightarrow \,\,\dfrac{1}{x_1}+\dfrac{1}{x_2} = \dfrac{x_1+x_2-2}{2}\,\,\Leftrightarrow \\ \\ \Leftrightarrow \,\,\dfrac{1}{x_1}+\dfrac{1}{x_2} =\dfrac{-\dfrac{-1}{1}-2}{2} \,\, \Leftrightarrow\,\,\boxed{\dfrac{1}{x_1}+\dfrac{1}{x_2} = -\dfrac{1}{2}}$

Answer 2

1.

log_5 (3x+4)=2 ⇒ 3x+4=5² ⇒ 3x=25-4 ⇒ x=21:3=7

conditiile  de existenta: 3x+4>0 ⇒ x>-4/3

S={7}

2.

$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1\cdot x_2}=\frac{\frac{-b}{a}}{\frac{c}{a}}=\frac{\frac{1}{1}}{\frac{-2}{1}}=\frac{1}{-2}$