Question

Resultado de (x-3) . (2x-1)>4?

Answer 1

x-3 + 1 = 4

2x 1. 5

5. (x-3) + 10 . 1 = 2.4

10x

5x-15+10×=8

5×-10×=8+15

15×=23

×=23

15

Explicação passo-a-passo:

Espero ter ajudado:)

Answer 2

$\boxed{\begin{array}{l}\sf (x-3)\cdot(2x-1)>4\\\sf 2x^2-x-6x+3>4\\\sf 2x^2-7x+3-4>0\\\sf 2x^2-7x-1>0\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm fac_{\!\!,}a}\\\sf f(x)=2x^2-7x-1\\\sf queremos~os~valores~de~x~tais~que~f(x)>0\\\underline{\rm c\acute alculo~das~ra\acute izes:}\\\sf 2x^2-7x-1=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-7)^2-4\cdot2\cdot(-1)\\\sf\Delta=49+8\\\sf\Delta=57\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-7)\pm\sqrt{57}}{2\cdot 2}\\\sf x=\dfrac{7\pm\sqrt{57}}{4}\begin{cases}\sf x_1=\dfrac{7+\sqrt{57}}{4}\\\sf x_2=\dfrac{7-\sqrt{57}}{4}\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Estudo~do~sinal\!:} \\\sf f(x)>0~para~x<\dfrac{7-\sqrt{57}}{4}~ou~x>\dfrac{7+\sqrt{57}}{4}\\\sf S=\bigg\{x\in\mathbb{R}/x<\dfrac{7-\sqrt{57}}{4}~ou~x>\dfrac{7+\sqrt{57}}{4}\bigg\}\end{array}}$