\(3+\left(x-1\right)^2=19\)
\(\left(x+1\right)^2=19-3\)
\(\left(x+1\right)^2=16\)
\(\left(x+1\right)^2=4^2\)
\(\Rightarrow x+1=4\)
\(x=4-1\)
\(x=3\)
Trả lời:
<=>x2-5/3 x-18=0
<=> 3x2-5x-54=0
Delta=25+648=673
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)=
\(x = {5 \pm \sqrt{673} \over 6}\)
Sorry, mình đọc nhầm đề
3+(x+1)^2=19
=>(x-1)^2=16
=> x-1={4,-4}
=> x={-3; 5}