Question

(x²+x+1)(x-1)-0,5(2x-1)(2x-1) решите пожалуйста

(3x-5)²-(1-2x)²=0 решите пожалуйста

Answer 1

$( {x}^{2} + x + 1)(x - 1) - 0.5(2x - 1)(2x - 1) = \\ = {x}^{3} - 1 - 0.5(2x - 1)(2x - 1) = \\ = {x}^{3} - 1 - 0.5(2x - 1)^{2} = {x}^{3} - 1 - 0.5(4 {x}^{2} - 4x + 1) = \\ = {x}^{3} - 1 - 2x ^{2} - 2x - 0.5 = {x}^{3} - 2 {x}^{2} - 2x - 1.5$

$(3x - 5) ^{2} - {(1 - 2x)}^{2} = 0 \\ 9 {x}^{2} - 30x + 25 - (1 - 4x + 4 {x}^{2} ) = 0 \\ 9 {x}^{2} - 30x + 25 -1 + 4x - 4 {x}^{2} = 0 \\ 5 {x}^{2} - 26x + 24 = 0 \\ x = \frac{ - ( - 26) + - \sqrt{( - 26) ^{2} - 4\times 5 \times 24} }{2 \times 5} \\ x = \frac{26 + - \sqrt{676 - 480} }{10} \\ x = \frac{26 + - \sqrt{196} }{10} \\ x = \frac{26 + - 14}{10} \\ \\ 1)x = \frac{26 + 14}{10} \\ x = \frac{40}{10} \\ x = 4 \\ \\ 2)x = \frac{26 - 14}{10} \\ x = \frac{12}{10} \\ x = 1.2$

Ответ: х1 = 1,2 ; х2 = 4