Question

Giải phương trình:

a) \(\sqrt{3x+4}-\sqrt{2x+1}=1\)

b) \(\sqrt{2x-1}-2\sqrt{x-1}=-1\)

Answer 1

\(a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{3x+4}=\sqrt{2x+1}+1\\ \Leftrightarrow3x+4=2x+2+2\sqrt{2x+1}\\ \Leftrightarrow x+2=2\sqrt{2x+1}\\ \Leftrightarrow x^2+4x+4=8x+4\\ \Leftrightarrow x^2-4x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\\ b,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{2x-1}=2\sqrt{x-1}-1\\ \Leftrightarrow2x-1=4x-3-4\sqrt{x-1}\\ \Leftrightarrow2x-2-4\sqrt{x-1}=0\\ \Leftrightarrow x-1-2\sqrt{x-1}=0\\ \Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)