Bài 1:
\(=\dfrac{\left(\dfrac{1}{4}y-x\right)\left(\dfrac{1}{16}y^2+\dfrac{1}{4}xy+x^2\right)}{\dfrac{1}{4}y-x}=\dfrac{1}{16}y^2+\dfrac{1}{4}xy+x^2\)
Câu 2:
\(\Leftrightarrow x-3+\left(2x+1\right)^2=20\)
\(\Leftrightarrow4x^2+4x+1+x-3=20\)
\(\Leftrightarrow4x^2+5x+2-20=0\)
=>4x^2+5x-18=0
hay \(x\in\left\{\dfrac{-5+\sqrt{313}}{8};\dfrac{-5-\sqrt{313}}{8}\right\}\)