a: \(\Leftrightarrow\dfrac{\left(x^2-2x+2\right)\left(x^2+x+1\right)-x^2\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow\dfrac{x^4+x^3+x^2-2x^3-2x^2-2x+2x^2+2x+2-x^4+x^3-x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow x\cdot2=3\)
=>x=3/2
b:
Sửa đề; \(\dfrac{x^2+2x}{\left(x+1\right)^2+3}-\dfrac{x^2-2x}{\left(x-1\right)^2+3}=\dfrac{16}{x^4+4x^2+16}\)
\(\Leftrightarrow\dfrac{x^2+2x}{x^2+2x+4}-\dfrac{x^2-2x}{x^2-2x+4}=\dfrac{16}{x^4+4x^2+16}\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-2x+4\right)-\left(x^2-2x\right)\left(x^2+2x+4\right)=16\)
\(\Leftrightarrow x^4-2x^3+4x^2+2x^3-4x^2+8x-\left(x^4+2x^3+4x^2-2x^3-4x^2-8x\right)=16\)
=>\(x^4+8x-x^4+8x=16\)
=>16x=16
=>x=1
