ĐKXĐ: \(x\ge0\)
Đặt \(\sqrt{x}+\sqrt{x+1}=a>0\)
\(\Rightarrow a^2=2x+1+2\sqrt{x\left(x+1\right)}\)
\(\Rightarrow2x+2\sqrt{x^2+x}=a^2-1\)
Phương trình trở thành:
\(a^2-1+a=1\)
\(\Leftrightarrow a^2+a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\sqrt{x+1}=1\)
\(\Leftrightarrow\sqrt{x}+\sqrt{x+1}-1=0\)
\(\Leftrightarrow\sqrt{x}+\dfrac{x}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow\sqrt{x}\left(1+\dfrac{\sqrt{x}}{\sqrt{x+1}+1}\right)=0\)
\(\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)
\(ĐK:x\ge0\)
Đặt \(\sqrt{x+1}+\sqrt{x}=a\left(a\ge0\right)\Leftrightarrow a^2=2x+1+2\sqrt{x^2+x}\)
\(PT\Leftrightarrow a^2+a-2=0\\ \Leftrightarrow\left(a-1\right)\left(a+2\right)=0\\ \Leftrightarrow a=1\left(a\ge0\right)\\ \Leftrightarrow\sqrt{x+1}+\sqrt{x}=1\\ \forall x\ge0\Leftrightarrow VT\ge\sqrt{0+1}+\sqrt{0}=1\)
Dấu \("="\Leftrightarrow x=0\left(tm\right)\)
Vậy PT có nghiệm \(x=0\)
