Question

Giải phương trình

a/ \(\sqrt{2x+1}\) =\(\dfrac{1}{x}\)

b/ \(\sqrt{x+1}\) = x+1

c/ \(\sqrt{x-1}\) = 1-x

d/ 2x + 3 + \(\dfrac{4}{x-1}\) = \(\dfrac{x^2+3}{x-1}\)

e/ \(\dfrac{x+1}{\sqrt{2x^2+1}}\) = 3x² + x +1

d/ x-\(\sqrt{3-x}\) = \(\sqrt{x-3}\) +3

e/ x² -\(\sqrt{2-x}\) = 3 + \(\sqrt{x-4}\)

f/ x² + \(\sqrt{-x-1}\) = 4 + \(\sqrt{-x-1}\)

Answer 1

b: ĐKXĐ: x>=-1

\(\sqrt{x+1}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+1\right)^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\cdot x=0\\x>=-1\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1\right\}\)

c: \(\sqrt{x-1}=1-x\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\1-x< =0\end{matrix}\right.\Leftrightarrow x=1\)

Do đó: x=1 là nghiệm của phương trình

d: \(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)(ĐKXĐ: x<>1)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)

\(\Leftrightarrow2x^2-2x+3x-3+4-x^2-3=0\)

\(\Leftrightarrow x^2+x-2=0\)

=>(x+2)(x-1)=0

=>x=-2(nhận) hoặc x=1(loại)