Question

помогите с 1 Б ;с 2 заданием; с 3 заданием; СРОЧНО СРОЧНО ДАЮ 10БАЛОВ

Answer 1

Объяснение:

1. b)

$m^2+4m-n^2-2n+3=m^2+4m+4-n^2-2n-1=(m+2)^2-(n^2+2n+1)=\\=(m+2)^2-(n+1)^2=(m+2+n+1)*(m+2-n-1)=(m+n+3)*(m-n+1).$

2.$a)\ \frac{a^2-a-20}{a^2-4a-5} =\frac{a^2-5a+4a-20}{a^2-5a+a-5}=\frac{a*(a-5)+4*(a-5)}{a*(a-5)+(a-5)} =\frac{(a-5)*(a+4)}{(a-5)*(a+1)} =\frac{a+4}{a+1} . \\b)\ \frac{x^3-2x^2+4x-3}{x^2-7x+6}=\frac{x^3-1-2x^2+4x-2}{x^2-x-6x+6}=\frac{(x-1)*(x^2+x+1)-2*(x^2-2x+1)}{x*(x-1)-6*(x-1)} =\frac{(x-1)*(x^2+x+1)-2*(x-1)^2}{(x-1)*(x-6)} =\\=\frac{(x-1)*(x^2+x+1-2*(x-1))}{(x-1)*(x-6)}=\frac{x^2+x+1-2x+2}{x-6}=\frac{x^2-x+3}{x-6}.$3.

$(x-2)^4-x^2+4x-4=0\\(x-2)^4-(x^2-4x+4)=0\\(x-2)^4-(x-2)^2=0\\(x-2)^2*((x-2)^2-1)=0\\(x-2)^2=0\\x-2=0\\x_1=2.\\x^2-4x+4-1=0\\x^2-4x+3=0\\D=4 \ \ \ \ \sqrt{D}=2\\x_2=1\ \ \ \ x_3=3.$

Ответ: x₁=2   x₂=1   x₃=3.