\(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
Đặt: \(\left\{{}\begin{matrix}a=\sqrt{3x^2+5x+8}\\b=\sqrt{3x^2+5x+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=3x^2+5x+8\\b^2=3x^2+5x+1\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}a-b=1\\a^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\\left(1+b\right)^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\1+2b+b^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x^2+5x+8}=4\\\sqrt{3x^2+5x+1}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+5x+8=16\\3x^2+5x+1=9\end{matrix}\right.\) \(\Leftrightarrow3x^2+5x-8=16\) \(\Leftrightarrow3x^2-3x+8x-8=0\)
\(\Leftrightarrow3x\left(x-1\right)+8\left(x-1\right)=0\) \(\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(n\right)\\x=-\dfrac{8}{3}\left(n\right)\end{matrix}\right.\)
\(\sqrt{3x^2+5x+8}-\sqrt{3x^2+6x+1}=1\)
Đặt : \(3x^2+5x+8=a\) . Phương trình trở thành :
\(\sqrt{a}-\sqrt{a-7}=1\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{a-7}\right)^2=1\)
\(\Leftrightarrow a-2\sqrt{a\left(a-7\right)}+a-7=1\)
\(\Leftrightarrow2a-2\sqrt{a\left(a-7\right)}=8\)
\(\Leftrightarrow2\sqrt{a\left(a-7\right)}=2a-8\)
\(\Leftrightarrow\sqrt{a\left(a-7\right)}=a-4\)
\(\Leftrightarrow a\left(a-7\right)=\left(a-4\right)^2\)
\(\Leftrightarrow a^2-7a=a^2-8a+16\)
\(\Leftrightarrow a=16\)
\(\Leftrightarrow3x^2+5x+8=16\)
\(\Leftrightarrow3x^2+5x-8=0\)
\(\Delta=5^2+4.3.8=25+96=121>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{121}}{6}=1\\x_2=\dfrac{-5-\sqrt{121}}{6}=-\dfrac{8}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{8}{3};1\right\}\)
