6/(y-1)(y-7)=2/(y-1)(y-7)+4/(y-1)(y-7)
4*(х-1)=2х-3
4х-2х=4-3
2хъ=1
х=1/2
ответ: да, есть 1/2
<span>2*x^3-3*x^2-4 = (2*x^2 + x + 2)(x - 2) = 0</span>
1)a^2-ac+2ab+b^2-bc=a^2+2ab+b^2-ac-bc=(a=b)^2-c(a+b)=(a+b)(b+a-c)
2)ab-c^2+ac+b^2=a(b+c)+(b-c)(b+c)=(b+c)(b-c+a)
3)(a+b)(b+a-c)\(b+c)(b-c+a)=1
-b³ · b²=-b³⁺²=-b⁵
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