Question

30PNTS
função exponencial
(0,01)^x+1=(0,001)^x2-x
helpa aew bro

Answer 1

$(0,01)^{x+1} = (0,001)^{x^2 - x} \\ ({10}^{-2})^{x+1} = ({10}^{-3})^{x^2-x} \\ {10}^{-2x-2} = {10}^{-3x^2+3x} \\ -2x-2=-3 x^2+3x \\ 3 x^2 -3x -2x -2 = 0 \\ \\ 3 x^2 -5x -2 = 0 \\ \Delta = b^2 - 4ac = {(-5)}^{2} - 4 \cdot 3 \cdot (-2) \\ \Delta = 25 + 24 \\ \Delta = 49$

$x = \frac{ b^2 - \sqrt{\Delta}}{2a} \\ \\ x_1 = \frac{ -(-5) + \sqrt{49} }{2 \cdot 3} = \frac{ 5 + 7 }{ 6 } = \frac{ 12 }{ 6 } \\ x_1 = 2 \\ \\ x_2 = \frac{ -(-5) - \sqrt{49} }{2 \cdot 3} = \frac{ 5 - 7 }{ 6 } = \frac{ -2 }{ 6 } \\ x_2 = - \frac{1}{3} \\ \\ \boxed{ \textrm{Resposta: S} = \left \{ 2; - \frac{1}{3} \right \} }$

Answer 2

Se eu entendi bem..

(0,01)ˣ⁺¹ = (0,001)ˣ²-ˣ

(1.10⁻²)ˣ⁺¹ = (10⁻³)ˣ²-ˣ

10⁻²ˣ⁻² = 10⁻³ˣ²⁺³ˣ

igualamos expoente...

-2x-2 = -3x²+ 3x

3x²-2x-3x-2 = 0

3x² -5x - 2 = 0

Δ = 25 +4*3*2

Δ = 25+24

Δ = 49

x' = 5 + 7/6

x' = 2

x'' = 5-7/6

x'' = -1/3