`9x^2-6x-5=\sqrt{3x+5}`
`đkxđ:x>=-5/3`
`pt<=>9x^2-12x+6x-8=\sqrt{3x+5}-3`
`<=>3x(3x-4)+2(3x-4)=(3x-4)/(\sqrt{3x+5}+3)`
`<=>(3x-4)(3x+2-1/(\sqrt{3x+5}+3))=0`
`<=>3x-4=0` vì `3x+2-1/(\sqrt{3x+5}+3) ne 0`
`<=>x=4/3`
Vậy `S={4/3}`
ĐKXĐ: ...
Đặt \(\sqrt{3x+5}=3y-1\) (1) ta được:
\(\left\{{}\begin{matrix}9x^2-6x-5=3y-1\\3x+5=\left(3y-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-6x-3y-4=0\\9y^2-6y-3x-4=0\end{matrix}\right.\)
Trừ vế cho vế:
\(\left(3x+3y\right)\left(x-y\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x+3y-1\right)=0\Rightarrow\left[{}\begin{matrix}x=y\\3x+3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3y-1=3x-1\\3y-1=-3x\end{matrix}\right.\) thế lên (1):
\(\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\\3x+5=9x^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{1}{3}< \dfrac{1}{3}\left(loại\right)\\x=\dfrac{1+\sqrt{21}}{6}>0\left(loại\right)\\x=\dfrac{1-\sqrt{21}}{6}\end{matrix}\right.\)
