Question

me ajundem pf??
3(x+x)-5x(10x-1)=33+x(x+1)-59

Answer 1

Resposta:

3(x+x)-5x(10x-1)=33+x(x+1)-59

15 x - 10x + 5 = 3x - 3

5x + 5 = 3x - 3  

2x=- 8  

x = - 4

Answer 2

Resposta:

Solução:

$\sf \displaystyle 3(x+x)-5x(10x-1)=33+x(x+1)-59$

$\sf \displaystyle 3x + 3x - 50x^{2} +5x = 33 +x^{2} +x - 59$

$\sf \displaystyle -50x^{2} -x^{2} +3x +3x+ 5x-x- 33 + 59 = 0$

$\sf \displaystyle -51x^{2}+ 6x+ 4x + 26 = 0$

$\sf \displaystyle -51x^{2}+ 10x + 26 = 0$

Determinar o Δ:

$\sf \displaystyle \Delta = b^2 -\:4ac$

$\sf \displaystyle \Delta =(10)^2 -\:4 \cdot (-51) \cdot 26$

$\sf \displaystyle \Delta = 100 + 5304$

$\sf \displaystyle \Delta = 5404$

Determinar as raízes da equação:

$\sf \displaystyle x = \dfrac{-\,b \pm \sqrt{ \Delta } }{2a} = \dfrac{-\,10 \pm \sqrt{ 5404 } }{2\cdot (-51)} = \dfrac{-\,10 \pm \sqrt{ 4 \cdot 1351 } }{- 2 \cdot 51} = \dfrac{10 \pm \sqrt{ 4 } \cdot \sqrt{1351} }{2 \cdot 51}$

$\sf \displaystyle x = \dfrac{10 \pm 2 \cdot \sqrt{1351} }{2 \cdot 51} = \dfrac{\diagup{\!\!\!2} \cdot \big [ \:5\pm 1 \cdot \sqrt{ 1351 }\; \big] }{ \diagup{\!\!\!2} \cdot 51} = \dfrac{ 5\pm \sqrt{ 1351 } }{ 51} \Rightarrow\begin{cases} \sf x_1 = &\sf \dfrac{5 + \sqrt{1351} }{51} \\\\ \sf x_2 = &\sf \dfrac{5 - \sqrt{1351} }{51}\end{cases}$

$\sf \boldsymbol{ \sf \displaystyle S = \Bigg\{ x \in \mathbb{R} \mid x = \dfrac{5- \sqrt{1351} }{51} \mbox{\sf \;e } x = \dfrac{5+ \sqrt{1351} }{51} \Bigg\} }$

Explicação passo-a-passo: