a) ĐKXĐ: \(x\ne\pm a\)
\(\frac{x-a}{x+a}-\frac{x+a}{x-a}+\frac{3a^2+a}{x^2-a^2}=0\)
\(\Leftrightarrow\frac{\left(x-a\right)^2-\left(x+a\right)^2+3a^2+a}{\left(x-a\right)\left(x+a\right)}=0\)
\(\Leftrightarrow x^2-2ax+a^2-x^2-2ax-a^2+3a^2+a=0\)
\(\Leftrightarrow-4ax+3a^2+a=0\)
Thay a = -3 vào phương trình, ta được :
\(-12x+27-3=0\)
\(\Leftrightarrow-12x+24=0\)
\(\Leftrightarrow-12x=-24\)
\(\Leftrightarrow x=2\)
Vậy khi \(a=-3\Leftrightarrow x=2\)
b) Thay a = 1 vào phương trình, ta được :
\(-4x+3+1=0\)
\(\Leftrightarrow-4x+4=0\)
\(\Leftrightarrow-4x=-4\)
\(\Leftrightarrow x=1\) (ktm đkxđ)
Vậy khi \(a=1\Leftrightarrow x\in\varnothing\)
c) Thay x = 0,5 vào phương trình :
\(-2a+3a^2+a=0\)
\(\Leftrightarrow3a^2-a=0\)
\(\Leftrightarrow a\left(3a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\3a-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=\frac{1}{3}\end{matrix}\right.\)
Vậy khi \(x=0,5\Leftrightarrow a\in\left\{0;\frac{1}{3}\right\}\)
\(\frac{x-a}{x+a}-\frac{x+a}{x-a}+\frac{3a^2+a}{x^2-a^2}=0\)
\(\Leftrightarrow\frac{\left(x-a\right)^2}{x^2-a^2}-\frac{\left(x+a\right)^2}{x^2-a^2}+\frac{3a^2+a}{x^2-a^2}\)
\(\Leftrightarrow\frac{x^2-2ax+a^2-x^2-2ax-a^2+3a^2+a}{x^2-a^2}\)
\(\Leftrightarrow\frac{-4ax+3a^2+a}{x^2-a^2}\)
a) Thay a=-3 vào phương trình:
\(\frac{\left(-4\cdot-3\cdot a\right)-\left(3\cdot-3^2\right)+\left(-3\right)}{x^2-\left(-3\right)^2}\Leftrightarrow\frac{12x+27-3}{x^2-9}\)
b)Thay a=1 vào phương trình:
\(\frac{\left(-4\cdot1\cdot x\right)+\left(3\cdot1^2\right)+1}{x^2-1^2}\Leftrightarrow\frac{-4x+4}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(\Leftrightarrow\frac{-4\cdot\left(x-1\right)}{\left(x-1\right)\cdot\left(x+1\right)}\Leftrightarrow\frac{-4}{x+1}\)
c) Thay x=0.5 vào phương trình
\(\frac{\left(-4\cdot0.5\cdot a\right)+3a^2+a}{0.5^2-a^2}\Leftrightarrow\frac{-2a+3a^2+a}{0.25-a^2}\)
\(\frac{a\cdot\left(-2+3a+1\right)}{0.25-a^2}\Leftrightarrow\frac{a\cdot\left(3a-1\right)}{0.25-a^2}\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\3a-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=0\\a=\frac{1}{3}\end{matrix}\right.\)
