Question

\(\sqrt{3x+19}-5=x\)

Answer 1

\(\sqrt{3x-19}-5=x\)

\(\sqrt{3x+19}=x+5\)

\(\left(\sqrt{3x+19}\right)^2=\left(x+5\right)^2\)

\(3x+19=x^2+10x+25\)

\(x^2+10x-3x=19-25\)

\(x^2+7x=-6\)

\(x^2+7x+6=0\)

\(\left(x+1\right)\left(x+6\right)=0\)

⇔ \(\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

Answer 2

\(\sqrt{3x+19}-5=x\) ĐKXĐ: 3x + 19 \(\ge\) 0

<=> \(\sqrt{3x+19}=5\)

<=> 3x + 19 = 25

<=> 3x = 6

<=> x = 2