Question

Разложите многочлены на множители:
$b^4 - b^2 + 4b + 4\x^2 - 4 - 3ax + 6a$
Решите уравнение:
$(3x-2)^2-(x+1)^2=0$

Answer 1

${b}^{4} - {b}^{2} + 4b + 4 = {b}^{2} ( {b}^{2} - 1) + 4(b + 1) = \ = {b}^{2} (b - 1)(b + 1) + 4(b + 1) = (b + 1)( {b}^{2} (b - 1) + 4) = \ = (b + 1)( {b}^{3} - {b}^{2} + 4)$

${x}^{2} - 4 - 3ax + 6a = {x}^{2} - 4 + 6a - 3ax = \ = - (4 - {x}^{2} ) + 3a(2 - x) = - (2 - x)(2 + x) + 3a(2 - x) = \ = (2 - x)( - (2 + x) + 3a) = (2 - x)(3a - x - 2)$

${(3x - 2)}^{2} - {(x + 1)}^{2} = 0 \ 9 {x}^{2} - 12x + 4 - ({x}^{2} + 2x + 1) = 0 \ 9 {x}^{2} - 12x + 4 - {x}^{2} - 2x - 1 = 0 \ 8 {x}^{2} - 14x + 3 = 0 \ D = {14}^{2} - 4 times 8 times 3 = 100 = {10}^{2} \ x_{1} = frac{14 + 10}{16} = frac{3}{2} = 1.5 \ x_{2} = frac{14 - 10}{16} = frac{1}{4} = 0.25$