a) ĐK: \(x\geq 0\)
Ta có: \(\sqrt{x}+\sqrt{x+1}=1\Leftrightarrow \sqrt{x}+\sqrt{x+1}-1=0\)
\(\Leftrightarrow \sqrt{x}+\frac{(x+1)-1}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow \sqrt{x}+\frac{x}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow \sqrt{x}\left(1+\frac{\sqrt{x}}{\sqrt{x+1}+1}\right)=0\)
Thấy rằng \(1+\frac{\sqrt{x}}{\sqrt{x+1}+1}>0, \forall x\geq 0\Rightarrow 1+\frac{\sqrt{x}}{\sqrt{x+1}+1}\neq 0\)
Do đó \(\sqrt{x}=0\Rightarrow x=0\) (thỏa mãn)
b) ĐK: \(x\geq 1\)
Ta thấy với mọi \(x\geq 1\) thì:\(\left\{\begin{matrix} \sqrt{x+4}\geq \sqrt{1+4}>2 \\ \sqrt{x-1}\geq 0\end{matrix}\right.\)
\(\Rightarrow \sqrt{x+4}+\sqrt{x-1}>2\)
Do đó pt \(\sqrt{x+4}+\sqrt{x-1}=2\) vô nghiệm
c)
\(\sqrt{1-4x+4x^2}=|5|=5\)
\(\Leftrightarrow \sqrt{(2x-1)^2}=5\)
\(\Leftrightarrow |2x-1|=5\Rightarrow \left[\begin{matrix} 2x-1=5\\ 2x-1=-5\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=-2\end{matrix}\right.\)
d) ĐK: \(x\geq -\frac{3}{4}\) hoặc \(x< -1\)
Ta có \(\sqrt{\frac{4x+3}{x+1}}=3\Rightarrow \frac{4x+3}{x+1}=3^2=9\)
\(\Rightarrow 4x+3=9(x+1)\)
\(\Leftrightarrow 5x+6=0\Rightarrow x=-\frac{6}{5}\) (thỏa mãn)
Vậy.........
