Question

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Answer 1

Explicação passo a passo:

Olá Boa noite!

$\red{\boxed{\mathbf{QUESTAO \: 6}}}$

$\sf{(3^{\frac{x}{2}})^{x - 1} = (\dfrac{3}{9})^{x - 3}}$

$\sf{(3^{\frac{x\cdot(x - 1)}{2}}) = (\dfrac{9}{3}^{-1})^{x - 3}}$

$\sf{(3^{\frac{x^2 - x}{2}}) = (\dfrac{9}{3})^{-1\cdot(x - 3)}}$

$\sf{(3^{\frac{x^2 - x}{2}}) = (3)^{- x + 3}}$

$\sf{(\backslash \!\!\! 3^{\frac{x^2 - x}{2}}) = (\backslash \!\!\! 3)^{- x + 3}}$

$\sf{\dfrac{x^2 - x}{2} = - x + 3}$

$\sf{x^2 - x = 2\cdot(- x + 3)}$

$\sf{x^2 - x = - 2x + 6}$

$\sf{x^2 - x + 2x - 6 = 0}$

$\sf{x^2 + x - 6 = 0}$

$\sf{\Delta = b^2 - 4ac}$

$\sf{\Delta = 1^2 - 4\cdot1\cdot(-6)}$

$\sf{\Delta = 1 + 24}$

$\sf{\Delta = 25}$

$\sf{x = \dfrac{- b \pm \sqrt{\Delta}}{2a}}$

$\sf{x = \dfrac{- 1 \pm \sqrt{25}}{2\cdot1}}$

$\sf{x = \dfrac{- 1 \pm 5}{2}}$

$•~~\sf{x' = \dfrac{- 1 + 5}{2} = \dfrac{4}{2} = 2}$

$•~~\sf{x" = \dfrac{- 1 - 5}{2} = - \dfrac{6}{2} = - 3}$

$\boxed{\sf{S = \left\{- 3~~;~~2\right\}}}$

$\red{\boxed{\mathbf{QUESTAO \: 7}}}$

$\sf{(\dfrac{5}{2})^x = 0,16}$

$\sf{(\dfrac{5}{2})^x = \dfrac{4}{25}}$

$\sf{(\dfrac{5}{2})^x = (\dfrac{2}{5})^2}$

$\sf{(\dfrac{5}{2})^x = (\dfrac{5}{2})^{-2}}$

$\sf{(\dfrac{\backslash \!\!\! 5}{\backslash \!\!\! 2})^x = (\dfrac{\backslash \!\!\! 5}{\backslash \!\!\! 2})^{-2}}$

$\boxed{\sf{x = - 2}}$