Bài 1:
d) ĐKXĐ: \(x\ne0\)
\(x+\dfrac{2a\left|x+a\right|}{x}=\dfrac{a^2}{x}\)
\(\Leftrightarrow\dfrac{x^2}{x}+\dfrac{2a\left|x+a\right|}{x}=\dfrac{a^2}{x}\)
\(\Rightarrow x^2+2a\left|x+a\right|=a^2\) (*)
-Khi \(x\ge-a\) thì:
(*) \(\Leftrightarrow x^2+2a\left(x+a\right)=a^2\)
\(\Leftrightarrow x^2+2ax+2a^2-a^2=0\)
\(\Leftrightarrow x^2+2ax+a^2=0\)
\(\Leftrightarrow\left(x+a\right)^2=0\)
\(\Leftrightarrow x+a=0\)
\(\Leftrightarrow x=-a\left(nhận\right)\)
-Khi \(x< -a\) thì:
(*) \(\Leftrightarrow x^2-2a\left(x+a\right)=a^2\)
\(\Leftrightarrow x^2-2ax-2a^2-a^2=0\)
\(\Leftrightarrow x^2+2ax-3a^2=0\)
\(\Leftrightarrow x^2+2ax+a^2-4a^2=0\)
\(\Leftrightarrow\left(x+a\right)^2-4a^2=0\)
\(\Leftrightarrow\left(x+a+2a\right)\left(x+a-2a\right)=0\)
\(\Leftrightarrow\left(x+3a\right)\left(x-a\right)=0\)
\(\Leftrightarrow x=-3a\left(nhận\right)\) hay \(x=a\left(nhận\right)\)
-Vậy khi \(x\ne0\) thì phương trình có tập nghiệm \(S=\left\{a;-a;-3a\right\}\)
