Б [x(y-1)+(y-1)]/[y(x+2)-(x+2)]=(y-1)(x+1)/[(x+2)(y-1)]-(x+1)/(x+2) в [c(a-2b)-(a-2b)]/(a-2b)²=(a-2b)(c-1)/(a-2b)²=(c-1)/(a-2b) г x²+4x+3=(x+1)(x+3) x1+x2=-4 U x1*x2=3⇒x1=-3 U x2=-1 (x+1)(x+3)/(x+1)²=(x+3)/(x+1) а {x+y=5 {x-y=7 прибавим 2x=12 x=6 6+y=5 y=5-6 y=-1 (6;-1)
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Verified answer
Б[x(y-1)+(y-1)]/[y(x+2)-(x+2)]=(y-1)(x+1)/[(x+2)(y-1)]-(x+1)/(x+2)
в
[c(a-2b)-(a-2b)]/(a-2b)²=(a-2b)(c-1)/(a-2b)²=(c-1)/(a-2b)
г
x²+4x+3=(x+1)(x+3)
x1+x2=-4 U x1*x2=3⇒x1=-3 U x2=-1
(x+1)(x+3)/(x+1)²=(x+3)/(x+1)
а
{x+y=5
{x-y=7
прибавим
2x=12
x=6
6+y=5
y=5-6
y=-1
(6;-1)