a: \(\Leftrightarrow x^4+3x^3+2x^2-2x-3x^2-2x+2=0\)
\(\Leftrightarrow x^4+3x^3-x^2-4x+2=0\)
\(\Leftrightarrow\left(x^2+x-1\right)\left(x^2+2x-2\right)=0\)
hay \(x\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2};-1+\sqrt{3};-1-\sqrt{3}\right\}\)
b: Đặt \(x^2-3x+5=a\)
Pt sẽ là \(\dfrac{a}{a-x}-\dfrac{a-2x}{a-3x}=\dfrac{-1}{4}\)
\(\Leftrightarrow\dfrac{a^2-3ax-\left(a^2-ax-2ax+2x^2\right)}{\left(a-x\right)\left(a-3x\right)}=\dfrac{-1}{4}\)
\(\Leftrightarrow4\left(a^2-3ax-a^2+3ax-2x^2\right)=-\left(a-x\right)\left(a-3x\right)\)
\(\Leftrightarrow4\cdot\left(-2x^2\right)=-\left(x^2-3x+5-x\right)\left(x^2-3x+5-3x\right)\)
\(\Leftrightarrow-8x^2=-\left(x^2-4x+5\right)\left(x^2-6x+5\right)\)
\(\Leftrightarrow\left(x^2-4x+5\right)\left(x^2-6x+5\right)=8x^2\)
\(\Leftrightarrow\left(x^2+5\right)^2-10x\left(x^2+5\right)+24x^2-8x^2=0\)
\(\Leftrightarrow\left(x^2+5\right)^2-10x\left(x^2+5\right)+16x^2=0\)
\(\Leftrightarrow\left(x^2-8x+5\right)=0\)
hay \(x\in\left\{4+\sqrt{11};4-\sqrt{11}\right\}\)
