a) PT \(\Leftrightarrow x^3+1-x^3+2x-9=0\)
\(\Leftrightarrow2x=8\) \(\Leftrightarrow x=4\)
Vậy \(x=4\)
b) PT \(\Leftrightarrow\left(x+4\right)^2-\left(x^2-16\right)=0\)
\(\Leftrightarrow8\cdot\left(x+4\right)=0\) \(\Leftrightarrow x=-4\)
Vậy \(x=-4\)
a)
(x+1).(x2-x+1)-x3+2x-9=0
(x+1).(x2-x.1+12)-x3+2x-9=0
x3+1-x3+2x-9 =0
(x3-x3)+2x+(1-9) =0
2x+8 =0
2.(x+4) =0
x+4 =0
x =-4
b)
(x+4)2-x2+16 =0
(x+4)2-(x2-16) =0
(x+4)2-(x2-42) =0
(x+4)(x+4)-(x-4)(x+4)=0
(x+4)((x+4)-(x-4)) =0
(x+4)(x+4-x+4) =0
(x+4)8 =0
x+4 =0
x =-4