ĐKXĐ\(x\ge\dfrac{-5}{3}\)
Đặt \(\sqrt{x^2+1}\) = a(a\(\ge\)0) => \(a^2=x^2+1\)
\(\sqrt[]{3x+5}\)= b(b\(\ge\)0) =>\(b^2=3x+5\)
PT đã cho trở thành: a+b=\(\sqrt{x^2+1+2\left(3x+5\right)}\)
<=> a+b=\(\sqrt{a^2+2b^2}\)
<=> \(a^2+2ab+b^2=a^2+2b^2\)
,<=> b(2a-b)=0
\(\left[{}\begin{matrix}b=0\\2a=b\end{matrix}\right.\)
+) Với b=0 <=>\(\sqrt{3x+5}=0\)
<=> x=\(\dfrac{-5}{3}\left(tm\right)\)
+) Với 2a=b <=>2\(\sqrt{x^2+1}=\sqrt{3x+5}\)
<=>\(4\left(x^2+1\right)=3x+5\)
<=>\(4x^2-3x-1=0\)
<=>\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=\dfrac{-1}{4}\left(tm\right)\end{matrix}\right.\)
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