Question

giải pt:

a) \(\left(\sqrt{5}+2\right)^{x-1}=\left(\sqrt{5}-2\right)^{\dfrac{x-1}{x+1}}\)

b) \(log_{x^2+3x}\left(x+3\right)-1=0\)

Answer 1

a.

ĐKXĐ: ...

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{5}-2}\right)^{x-1}=\left(\sqrt{5}-2\right)^{\dfrac{x-1}{x+1}}\)

\(\Leftrightarrow\left(\sqrt{5}-2\right)^{1-x}=\left(\sqrt{5}-2\right)^{\dfrac{x-1}{x+1}}\)

\(\Leftrightarrow1-x=\dfrac{x-1}{x+1}\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x+3>0\\x^2+3x>0\end{matrix}\right.\) \(\Rightarrow x>3\)

\(log_{x^2+3x}\left(x+3\right)=1\)

\(\Rightarrow x+3=x^2+3x\)

\(\Rightarrow x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\left(loại\right)\end{matrix}\right.\)