Решите уравнения1) x^2-5= (x+5)(2x-1)2) (2x+3)(3x+1)=11x+303)2x-(x+1)^2=3x^2-64) 6a^2-(a+2)^2= - 4(a-4)5)x(7-6x)=(1-3x)(1+2x)6)(5y+2)(y-3)= - 13(2+y)

x^2-5= (x+5)(2x-1)x^2 - 5 = 2x^2 - x + 10x - 52x^2 - x^2 - x - 10x - 5 + 5 = 0x^2 - 9x = 0x * (x - 9) = 0x = 0 или x = 9(2x+3)(3x+1)=11x+306x^2 + 2x + 9x + 3 = 11x + 306x^2 + 11x - 11x = 30 - 36x^2 = 272x^2 = 9x^2 = 9/2x = 3/ √ 22x-(x+1)^2=3x^2-62x - x^2 - 1 - 2x = 3x^2 - 63x^2 + x^2 = 54x^2 = 5x^2 = 5/4x = √5/26a^2-(a+2)^2= - 4(a-4)6a^2 - a^2 - 4 - 4a = - 4a + 165a^2 = 20a^2 = 4a = 2x(7-6x)=(1-3x)(1+2x)7x - 6x^2 = 1 + 2x - 3x - 6x^27x - 2x + 3x = 18x = 1x = 1/8(5y+2)(y-3)= - 13(2+y)5y^2 - 15y + 2y - 6 = - 26 - 13y5y^2 - 13y + 13y = - 26 + 65y^2 = - 20y^2 = - 4y = решений нет, поскольку из отрицательного числа квадратный корень извлечь невозможно.