a.
ĐKXĐ: \(x\ge-1\)
\(\sqrt{x+1}+\sqrt{2x+3}=5\)
\(\Leftrightarrow\sqrt{x+1}-2+\sqrt{2x+3}-3=0\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{x+1}+2}+\dfrac{2\left(x-3\right)}{\sqrt{2x+3}+3}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x+1}+2}+\dfrac{2}{\sqrt{2x+3}+3}\right)=0\)
\(\Leftrightarrow x-3=0\) (do \(\dfrac{1}{\sqrt{x+1}+2}+\dfrac{2}{\sqrt{2x+3}+3}>0\))
\(\Leftrightarrow x=3\)
b.
Dạng câu này có cách làm riêng, ko tìm ĐKXĐ:
\(\sqrt{3x^2+4x+1}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\3x^2+4x+1=\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\2x^2+6x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
\(a,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{x+1}-2+\sqrt{2x+3}-3=0\\ \Leftrightarrow\dfrac{x-3}{\sqrt{x+1}+2}+\dfrac{2\left(x-3\right)}{\sqrt{2x+3}+3}=0\\ \Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x+1}+2}+\dfrac{2}{\sqrt{2x+3}+3}\right)=0\)
Dễ thấy ngoặc lớn luôn >0
Do đó \(x-3=0\Leftrightarrow x=3\)
\(b,ĐK:x\le-1\\ PT\Leftrightarrow\sqrt{3x^2+4x+1}=x-1\\ \Leftrightarrow3x^2+4x+1=x^2-2x+1\\ \Leftrightarrow2x^2+6x=0\\ \Leftrightarrow2x\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
