Question

Mă poate ajuta cineva cu aceasta problemă? Dacă nu știți nu e nevoie sa răspundeți.​

Answer 1

$\displaystyle\it\\b)~xy+5x=-13 \Leftrightarrow x(y+5)=-13,~\underline{casework}~:\\13=-13\cdot1=13(-1)=1(-13)=-1\cdot13.\\\\d)~xy+x+y=8 | +1 \implies xy+x+y+1=x(y+1)+(y+1)=\\(x+1)(y+1)=9,~\underline{casework}~: 9=1\cdot9=(-1)(-9)=\\9\cdot1=(-9)(-1)=3\cdot3=(-3)(-3).\\\\e)~x^2+x-6=x^2+2x-3x-6=x(x+2)-3(x+2)=\\(x-3)(x+2)=0 \implies x\in\left\{-2,3\right\}.\\\\f)~x^2-x-12=x^2+3x-4x-12=x(x+3)-4(x+3)=\\(x-4)(x+3)=0 \implies x\in\left\{-3,4\right\}.$

$\displaystyle\it\\g)~x^2-6x=16 \Leftrightarrow x^2-6x-16=x^2+2x-8x-16=\\x(x+2)-8(x+2)=(x-8)(x+2)=0 \implies x\in\left\{-2,8\right\}.$