a)
ĐKXĐ: x ≠ 2 và x ≠ -2
b)
\(\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}=0\)
<=> \(\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=0\)
=>\(x^3-x\left(x+2\right)-2\left(x-2\right)=0\)
<=> \(x^3-x^2-4x+4=0\)
<=> \(x^2\left(x-1\right)-4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2-4\right)=0\)
<=> (x - 1 )(x - 2)(x + 2)=0
Vậy x - 1 =0 hoặc x - 2 =0 hoặc x + 2 =0
1) x - 1 =0
<=> x =1 ( thỏa mãn ĐKXĐ)
2) x - 2 =0
<=> x =2 ( không thỏa mãn ĐKXĐ)
3) x + 2=0
<=> x = -2 ( không thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm x = 1
a, DKXD:\(x\ne2,x\ne-2\)
C=\(\dfrac{x^3}{x^2-4}-\dfrac{x\left(x+2\right)}{x^2-4}-\dfrac{2\left(x-2\right)}{x^2-4}\)
\(\Rightarrow x^3-x^2-2x-2x+4=x^3-x^2-4x+4\)\(=x^2\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\)
Để C=0 \(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
